\displaystyle \textbf{Question 1: } \text{From a group of }15\text{ cricket players, a team of }11\text{ players is to be}
\displaystyle \text{chosen. In how many ways can this be done?}
\displaystyle \text{Answer:}
\displaystyle \text{A team of }11\text{ players can be chosen from }15\text{ players in}
\displaystyle {}^{15}\mathrm{C}_{11}\text{ ways.}
\displaystyle {}^{15}\mathrm{C}_{11}=\frac{15!}{11!\,4!}
\displaystyle =\frac{15\times14\times13\times12}{4\times3\times2\times1}
\displaystyle =1365
\displaystyle \therefore \text{The required number of ways is }1365.
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{How many different boat parties of }8,\text{ consisting of }5\text{ boys and}
\displaystyle \text{3 girls, can be made from }25\text{ boys and }10\text{ girls?}
\displaystyle \text{Answer:}
\displaystyle \text{The }5\text{ boys can be chosen from }25\text{ boys in }{}^{25}\mathrm{C}_5\text{ ways.}
\displaystyle \text{The }3\text{ girls can be chosen from }10\text{ girls in }{}^{10}\mathrm{C}_3\text{ ways.}
\displaystyle \therefore \text{Required number of boat parties}={}^{25}\mathrm{C}_5\times{}^{10}\mathrm{C}_3
\displaystyle =\frac{25!}{5!\,20!}\times\frac{10!}{3!\,7!}
\displaystyle =\frac{25\times24\times23\times22\times21}{5\times4\times3\times2\times1}
\displaystyle \qquad\times\frac{10\times9\times8}{3\times2\times1}
\displaystyle =6375600
\displaystyle \therefore \text{The required number of different boat parties is }6375600.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In how many ways can a student choose }5\text{ courses out of }9\text{ courses if}
\displaystyle \text{2 courses are compulsory for every student?}
\displaystyle \text{Answer:}
\displaystyle 2\text{ courses are compulsory for every student.}
\displaystyle \therefore \text{The remaining }3\text{ courses are to be chosen from the other }7\text{ courses.}
\displaystyle \text{Required number of ways}={}^{7}\mathrm{C}_{3}
\displaystyle =\frac{7!}{3!\,4!}
\displaystyle =\frac{7\times6\times5}{3\times2\times1}
\displaystyle =35
\displaystyle \therefore \text{The student can choose the courses in }35\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In how many ways can a football team of }11\text{ players be selected from}
\displaystyle 16\text{ players? How many of these will (i) include }2\text{ particular players?}
\displaystyle \text{(ii) exclude }2\text{ particular players?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of ways to select }11\text{ players from }16\text{ players}
\displaystyle ={}^{16}\mathrm{C}_{11}
\displaystyle =\frac{16!}{11!\,5!}
\displaystyle =\frac{16\times15\times14\times13\times12}{5\times4\times3\times2\times1}
\displaystyle =4368
\displaystyle \text{i) If the two particular players are included, they occupy }2\text{ places.}
\displaystyle \therefore \text{The remaining }9\text{ players are to be selected from the other }14\text{ players.}
\displaystyle \text{Required number of ways}={}^{14}\mathrm{C}_{9}
\displaystyle =\frac{14!}{9!\,5!}
\displaystyle =\frac{14\times13\times12\times11\times10}{5\times4\times3\times2\times1}
\displaystyle =2002
\displaystyle \text{ii) If the two particular players are excluded, the team is selected from the}
\displaystyle \text{remaining }14\text{ players.}
\displaystyle \text{Required number of ways}={}^{14}\mathrm{C}_{11}
\displaystyle =\frac{14!}{11!\,3!}
\displaystyle =\frac{14\times13\times12}{3\times2\times1}
\displaystyle =364
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{There are }10\text{ professors and }20\text{ students. A committee of }2
\displaystyle \text{professors and }3\text{ students is to be formed. Find the number of ways in which}
\displaystyle \text{this can be done. Further, find in how many of these committees:}
\displaystyle \text{(i) a particular professor is included.}
\displaystyle \text{(iii) a particular student is excluded.}
\displaystyle \text{(ii) a particular student is included.}
\displaystyle \text{Answer:}
\displaystyle \text{Number of ways to form the committee}
\displaystyle ={}^{10}\mathrm{C}_{2}\times{}^{20}\mathrm{C}_{3}
\displaystyle =\frac{10!}{2!\,8!}\times\frac{20!}{3!\,17!}
\displaystyle =\frac{10\times9}{2}\times\frac{20\times19\times18}{6}
\displaystyle =51300
\displaystyle \text{i) If a particular professor is included, the remaining professor is chosen from the other }9\text{ professors.}
\displaystyle \text{Required number of ways}={}^{9}\mathrm{C}_{1}\times{}^{20}\mathrm{C}_{3}
\displaystyle =\frac{9!}{1!\,8!}\times\frac{20!}{3!\,17!}
\displaystyle =9\times\frac{20\times19\times18}{6}
\displaystyle =10260
\displaystyle \text{iii) If a particular student is excluded, }3\text{ students are chosen from the remaining }19\text{ students.}
\displaystyle \text{Required number of ways}={}^{10}\mathrm{C}_{2}\times{}^{19}\mathrm{C}_{3}
\displaystyle =\frac{10!}{2!\,8!}\times\frac{19!}{3!\,16!}
\displaystyle =\frac{10\times9}{2}\times\frac{19\times18\times17}{6}
\displaystyle =43605
\displaystyle \text{ii) If a particular student is included, the remaining }2\text{ students are chosen from the other }19\text{ students.}
\displaystyle \text{Required number of ways}={}^{10}\mathrm{C}_{2}\times{}^{19}\mathrm{C}_{2}
\displaystyle =\frac{10!}{2!\,8!}\times\frac{19!}{2!\,17!}
\displaystyle =\frac{10\times9}{2}\times\frac{19\times18}{2}
\displaystyle =7695
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{How many different products can be obtained by multiplying two or more}
\displaystyle \text{of the numbers }3,5,7,11\text{ without repetition?}
\displaystyle \text{Answer:}
\displaystyle \text{Since }3,5,7\text{ and }11\text{ are distinct prime numbers, every different selection}
\displaystyle \text{of the numbers gives a different product.}
\displaystyle \text{The products may contain }2,3\text{ or }4\text{ of the given numbers.}
\displaystyle \therefore \text{Required number of different products}
\displaystyle ={}^{4}\mathrm{C}_{2}+{}^{4}\mathrm{C}_{3}+{}^{4}\mathrm{C}_{4}
\displaystyle =6+4+1
\displaystyle =11
\displaystyle \therefore \text{The required number of different products is }11.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{From a class of }12\text{ boys and }10\text{ girls, }10\text{ students are to be chosen}
\displaystyle \text{for a competition, including at least }4\text{ boys and }4\text{ girls. The }2\text{ girls who}
\displaystyle \text{won prizes last year must be included. In how many ways can the selection be made?}
\displaystyle \text{Answer:}
\displaystyle \text{The }2\text{ girls who won prizes last year are compulsory.}
\displaystyle \therefore \text{The remaining students are to be selected from }12\text{ boys and }8\text{ girls.}
\displaystyle \text{Since at least }4\text{ boys and }4\text{ girls are required, the possible selections are:}
\displaystyle 6\text{ boys and }4\text{ girls},\quad5\text{ boys and }5\text{ girls},\quad\text{or }4\text{ boys and }6\text{ girls}.
\displaystyle \text{As }2\text{ girls are already included, the required number of ways is}
\displaystyle {}^{12}\mathrm{C}_{6}\times{}^{8}\mathrm{C}_{2}+{}^{12}\mathrm{C}_{5}\times{}^{8}\mathrm{C}_{3}
\displaystyle \qquad+{}^{12}\mathrm{C}_{4}\times{}^{8}\mathrm{C}_{4}
\displaystyle =924\times28+792\times56+495\times70
\displaystyle =25872+44352+34650
\displaystyle =104874
\displaystyle \therefore \text{The selection can be made in }104874\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{How many different selections of }4\text{ books can be made from }10
\displaystyle \text{different books, if (i) there is no restriction; (ii) two particular books are}
\displaystyle \text{always selected; (iii) two particular books are never selected?}
\displaystyle \text{Answer:}
\displaystyle \text{i) If there is no restriction, the required number of selections}
\displaystyle ={}^{10}\mathrm{C}_{4}
\displaystyle =\frac{10!}{4!\,6!}
\displaystyle =\frac{10\times9\times8\times7}{4\times3\times2\times1}
\displaystyle =210
\displaystyle \text{ii) If two particular books are always selected, the remaining }2\text{ books are}
\displaystyle \text{chosen from the other }8\text{ books.}
\displaystyle \text{Required number of selections}={}^{8}\mathrm{C}_{2}
\displaystyle =\frac{8!}{2!\,6!}
\displaystyle =\frac{8\times7}{2\times1}
\displaystyle =28
\displaystyle \text{iii) If two particular books are never selected, all }4\text{ books are chosen from}
\displaystyle \text{the remaining }8\text{ books.}
\displaystyle \text{Required number of selections}={}^{8}\mathrm{C}_{4}
\displaystyle =\frac{8!}{4!\,4!}
\displaystyle =\frac{8\times7\times6\times5}{4\times3\times2\times1}
\displaystyle =70
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{From }4\text{ officers and }8\text{ soldiers, in how many ways can }6\text{ persons be}
\displaystyle \text{chosen (i) to include exactly one officer; (ii) to include at least one officer?}
\displaystyle \text{Answer:}
\displaystyle \text{i) To include exactly one officer, }1\text{ officer is chosen from }4\text{ officers and}
\displaystyle 5\text{ soldiers are chosen from }8\text{ soldiers.}
\displaystyle \text{Required number of ways}={}^{4}\mathrm{C}_{1}\times{}^{8}\mathrm{C}_{5}
\displaystyle =\frac{4!}{1!\,3!}\times\frac{8!}{5!\,3!}
\displaystyle =4\times\frac{8\times7\times6}{3\times2\times1}
\displaystyle =4\times56
\displaystyle =224
\displaystyle \text{ii) Total number of ways to choose }6\text{ persons from }12\text{ persons}
\displaystyle ={}^{12}\mathrm{C}_{6}
\displaystyle =\frac{12!}{6!\,6!}
\displaystyle =\frac{12\times11\times10\times9\times8\times7}{6\times5\times4\times3\times2\times1}
\displaystyle =924
\displaystyle \text{Number of selections containing no officer}={}^{8}\mathrm{C}_{6}
\displaystyle =\frac{8!}{6!\,2!}
\displaystyle =\frac{8\times7}{2\times1}
\displaystyle =28
\displaystyle \therefore \text{Number of selections containing at least one officer}
\displaystyle ={}^{12}\mathrm{C}_{6}-{}^{8}\mathrm{C}_{6}
\displaystyle =924-28
\displaystyle =896
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{A sports team of }11\text{ students is to be constituted, choosing at least }5
\displaystyle \text{students from Class XI and at least }5\text{ students from Class XII. If there are }20
\displaystyle \text{students in each class, in how many ways can the team be constituted?}
\displaystyle \text{Answer:}
\displaystyle \text{Since the team consists of }11\text{ students and at least }5\text{ students must be chosen}
\displaystyle \text{from each class, the possible selections are:}
\displaystyle 5\text{ students from Class XI and }6\text{ students from Class XII, or}
\displaystyle 6\text{ students from Class XI and }5\text{ students from Class XII.}
\displaystyle \therefore \text{Required number of ways}
\displaystyle ={}^{20}\mathrm{C}_{5}\times{}^{20}\mathrm{C}_{6}+{}^{20}\mathrm{C}_{6}\times{}^{20}\mathrm{C}_{5}
\displaystyle =2\times{}^{20}\mathrm{C}_{5}\times{}^{20}\mathrm{C}_{6}
\displaystyle =2\times\frac{20!}{5!\,15!}\times\frac{20!}{6!\,14!}
\displaystyle =2\times15504\times38760
\displaystyle =1201870080
\displaystyle \therefore \text{The team can be constituted in }1201870080\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{A student has to answer }10\text{ questions, choosing at least }4\text{ questions}
\displaystyle \text{from each of Part A and Part B. If there are }6\text{ questions in Part A and }7
\displaystyle \text{questions in Part B, in how many ways can the student choose }10\text{ questions?}
\displaystyle \text{Answer:}
\displaystyle \text{The student must choose at least }4\text{ questions from each part.}
\displaystyle \text{Therefore, the possible selections from Parts A and B respectively are}
\displaystyle (4,6),\quad(5,5)\quad\text{and}\quad(6,4).
\displaystyle \therefore \text{Required number of ways}
\displaystyle ={}^{6}\mathrm{C}_{4}\times{}^{7}\mathrm{C}_{6}+{}^{6}\mathrm{C}_{5}\times{}^{7}\mathrm{C}_{5}
\displaystyle \qquad+{}^{6}\mathrm{C}_{6}\times{}^{7}\mathrm{C}_{4}
\displaystyle =15\times7+6\times21+1\times35
\displaystyle =105+126+35
\displaystyle =266
\displaystyle \therefore \text{The student can choose the questions in }266\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In an examination, a student has to answer }4\text{ questions out of }5
\displaystyle \text{questions. Questions }1\text{ and }2\text{ are compulsory. Determine the number of ways}
\displaystyle \text{in which the student can make the choice.}
\displaystyle \text{Answer:}
\displaystyle \text{Questions }1\text{ and }2\text{ are compulsory.}
\displaystyle \therefore \text{The student must choose }2\text{ more questions from the remaining }3\text{ questions.}
\displaystyle \text{Required number of ways}={}^{3}\mathrm{C}_{2}
\displaystyle =\frac{3!}{2!\,1!}
\displaystyle =3
\displaystyle \therefore \text{The student can make the choice in }3\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{A candidate is required to answer }7\text{ questions out of }12\text{ questions,}
\displaystyle \text{which are divided into two groups, each containing }6\text{ questions. The candidate is}
\displaystyle \text{not permitted to attempt more than }5\text{ questions from either group. In how many}
\displaystyle \text{ways can the candidate choose the }7\text{ questions?}
\displaystyle \text{Answer:}
\displaystyle \text{Since no more than }5\text{ questions can be chosen from either group, the possible}
\displaystyle \text{selections from the two groups are }(2,5),(3,4),(4,3)\text{ and }(5,2).
\displaystyle \therefore \text{Required number of ways}
\displaystyle ={}^{6}\mathrm{C}_{2}\times{}^{6}\mathrm{C}_{5}+{}^{6}\mathrm{C}_{3}\times{}^{6}\mathrm{C}_{4}
\displaystyle \qquad+{}^{6}\mathrm{C}_{4}\times{}^{6}\mathrm{C}_{3}+{}^{6}\mathrm{C}_{5}\times{}^{6}\mathrm{C}_{2}
\displaystyle =15\times6+20\times15+15\times20+6\times15
\displaystyle =90+300+300+90
\displaystyle =780
\displaystyle \therefore \text{The candidate can choose the }7\text{ questions in }780\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{There are }10\text{ points in a plane, of which }4\text{ are collinear. How many}
\displaystyle \text{different straight lines can be drawn by joining these points?}
\displaystyle \text{Answer:}
\displaystyle \text{The number of lines determined by }10\text{ points, taking }2\text{ points at a time, is}
\displaystyle {}^{10}\mathrm{C}_{2}=\frac{10!}{2!\,8!}=\frac{10\times9}{2}=45
\displaystyle \text{The }4\text{ collinear points contribute}
\displaystyle {}^{4}\mathrm{C}_{2}=\frac{4!}{2!\,2!}=\frac{4\times3}{2}=6
\displaystyle \text{pairs, but all these pairs represent only one straight line.}
\displaystyle \therefore \text{Required number of straight lines}
\displaystyle =45-6+1=40
\displaystyle \therefore \text{The required number of straight lines is }40.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Find the number of diagonals of (i) a hexagon (ii) a polygon of }16\text{ sides.}
\displaystyle \text{Answer:}
\displaystyle \text{An }n\text{-sided polygon has }n\text{ vertices. Joining any two vertices gives either}
\displaystyle \text{a side or a diagonal.}
\displaystyle \text{i) Number of diagonals of a hexagon}
\displaystyle ={}^{6}\mathrm{C}_{2}-6
\displaystyle =\frac{6!}{2!\,4!}-6
\displaystyle =\frac{6\times5}{2}-6
\displaystyle =15-6=9
\displaystyle \text{ii) Number of diagonals of a }16\text{-sided polygon}
\displaystyle ={}^{16}\mathrm{C}_{2}-16
\displaystyle =\frac{16!}{2!\,14!}-16
\displaystyle =\frac{16\times15}{2}-16
\displaystyle =120-16=104
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{How many triangles can be obtained by joining }12\text{ points, five of which}
\displaystyle \text{are collinear?}
\displaystyle \text{Answer:}
\displaystyle \text{The total number of ways of choosing }3\text{ points from }12\text{ points is}
\displaystyle {}^{12}\mathrm{C}_{3}
\displaystyle \text{Among these, }{}^{5}\mathrm{C}_{3}\text{ selections are collinear and do not form triangles.}
\displaystyle \therefore \text{Required number of triangles}
\displaystyle ={}^{12}\mathrm{C}_{3}-{}^{5}\mathrm{C}_{3}
\displaystyle =\frac{12!}{3!\,9!}-\frac{5!}{3!\,2!}
\displaystyle =\frac{12\times11\times10}{3\times2\times1}-\frac{5\times4}{2\times1}
\displaystyle =220-10=210
\displaystyle \therefore \text{The required number of triangles is }210.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In how many ways can a committee of }5\text{ persons be formed from }6\text{ men}
\displaystyle \text{and }4\text{ women, if at least one woman has to be selected?}
\displaystyle \text{Answer:}
\displaystyle \text{The committee may contain }1,2,3\text{ or }4\text{ women.}
\displaystyle \therefore \text{Required number of ways}
\displaystyle ={}^{4}\mathrm{C}_{1}\times{}^{6}\mathrm{C}_{4}+{}^{4}\mathrm{C}_{2}\times{}^{6}\mathrm{C}_{3}
\displaystyle \qquad+{}^{4}\mathrm{C}_{3}\times{}^{6}\mathrm{C}_{2}+{}^{4}\mathrm{C}_{4}\times{}^{6}\mathrm{C}_{1}
\displaystyle =4\times15+6\times20+4\times15+1\times6
\displaystyle =60+120+60+6
\displaystyle =246
\displaystyle \therefore \text{The committee can be formed in }246\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In a village, there are }87\text{ families, of which }52\text{ families have at most}
\displaystyle 2\text{ children. In a rural development programme, }20\text{ families are to be chosen}
\displaystyle \text{for assistance, of which at least }18\text{ families must have at most }2\text{ children.}
\displaystyle \text{In how many ways can the choice be made?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of families having at most }2\text{ children}=52
\displaystyle \text{Number of families having more than }2\text{ children}=87-52=35
\displaystyle \text{Since at least }18\text{ of the }20\text{ families must have at most }2\text{ children,}
\displaystyle \text{the possible selections are }(18,2),(19,1)\text{ and }(20,0).
\displaystyle \therefore \text{Required number of ways}
\displaystyle ={}^{52}\mathrm{C}_{18}\times{}^{35}\mathrm{C}_{2}+{}^{52}\mathrm{C}_{19}\times{}^{35}\mathrm{C}_{1}
\displaystyle \qquad+{}^{52}\mathrm{C}_{20}\times{}^{35}\mathrm{C}_{0}
\displaystyle \therefore \text{The required number of ways is}
\displaystyle {}^{52}\mathrm{C}_{18}\times{}^{35}\mathrm{C}_{2}+{}^{52}\mathrm{C}_{19}\times{}^{35}\mathrm{C}_{1}+{}^{52}\mathrm{C}_{20}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{A group consists of }4\text{ girls and }7\text{ boys. In how many ways can a team}
\displaystyle \text{of }5\text{ members be selected if the team has (i) no girl; (ii) at least one boy and}
\displaystyle \text{one girl; (iii) at least }3\text{ girls?}
\displaystyle \text{Answer:}
\displaystyle \text{i) If the team has no girl, all }5\text{ members are chosen from the }7\text{ boys.}
\displaystyle \text{Required number of ways}={}^{7}\mathrm{C}_{5}
\displaystyle =\frac{7!}{5!\,2!}
\displaystyle =\frac{7\times6}{2\times1}
\displaystyle =21
\displaystyle \text{ii) If the team has at least one boy and one girl, the possible selections are}
\displaystyle (1\text{ girl},4\text{ boys}),(2\text{ girls},3\text{ boys}),(3\text{ girls},2\text{ boys})
\displaystyle \text{and }(4\text{ girls},1\text{ boy}).
\displaystyle \therefore \text{Required number of ways}
\displaystyle ={}^{4}\mathrm{C}_{1}\times{}^{7}\mathrm{C}_{4}+{}^{4}\mathrm{C}_{2}\times{}^{7}\mathrm{C}_{3}
\displaystyle \qquad+{}^{4}\mathrm{C}_{3}\times{}^{7}\mathrm{C}_{2}+{}^{4}\mathrm{C}_{4}\times{}^{7}\mathrm{C}_{1}
\displaystyle =4\times35+6\times35+4\times21+1\times7
\displaystyle =140+210+84+7
\displaystyle =441
\displaystyle \text{iii) If the team has at least }3\text{ girls, it may contain }3\text{ girls and }2\text{ boys,}
\displaystyle \text{or }4\text{ girls and }1\text{ boy.}
\displaystyle \therefore \text{Required number of ways}
\displaystyle ={}^{4}\mathrm{C}_{3}\times{}^{7}\mathrm{C}_{2}+{}^{4}\mathrm{C}_{4}\times{}^{7}\mathrm{C}_{1}
\displaystyle =4\times21+1\times7
\displaystyle =84+7
\displaystyle =91
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{A committee of }3\text{ persons is to be constituted from a group of }2\text{ men}
\displaystyle \text{and }3\text{ women. In how many ways can this be done? How many of these committees}
\displaystyle \text{would consist of }1\text{ man and }2\text{ women?}
\displaystyle \text{Answer:}
\displaystyle \text{i) Total number of persons}=2+3=5
\displaystyle \therefore \text{Number of ways to form a committee of }3\text{ persons}
\displaystyle ={}^{5}\mathrm{C}_{3}
\displaystyle =\frac{5!}{3!\,2!}
\displaystyle =\frac{5\times4}{2\times1}
\displaystyle =10
\displaystyle \text{ii) For a committee consisting of }1\text{ man and }2\text{ women,}
\displaystyle \text{Required number of ways}={}^{2}\mathrm{C}_{1}\times{}^{3}\mathrm{C}_{2}
\displaystyle =\frac{2!}{1!\,1!}\times\frac{3!}{2!\,1!}
\displaystyle =2\times3
\displaystyle =6
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Find the number of (i) diagonals (ii) triangles formed in a decagon.}
\displaystyle \text{Answer:}
\displaystyle \text{A decagon has }10\text{ vertices.}
\displaystyle \text{i) Number of diagonals}
\displaystyle ={}^{10}\mathrm{C}_{2}-10
\displaystyle =\frac{10!}{2!\,8!}-10
\displaystyle =\frac{10\times9}{2}-10
\displaystyle =45-10
\displaystyle =35
\displaystyle \text{ii) Number of triangles}
\displaystyle ={}^{10}\mathrm{C}_{3}
\displaystyle =\frac{10!}{3!\,7!}
\displaystyle =\frac{10\times9\times8}{3\times2\times1}
\displaystyle =120
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Determine the number of }5\text{-card combinations from a deck of }52\text{ cards,}
\displaystyle \text{if at least one of the }5\text{ cards has to be a king.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards}=52,\qquad\text{Number of kings}=4
\displaystyle \text{A }5\text{-card hand may contain }1,2,3\text{ or }4\text{ kings.}
\displaystyle \therefore \text{Required number of combinations}
\displaystyle ={}^{4}\mathrm{C}_{1}\times{}^{48}\mathrm{C}_{4}+{}^{4}\mathrm{C}_{2}\times{}^{48}\mathrm{C}_{3}
\displaystyle \qquad+{}^{4}\mathrm{C}_{3}\times{}^{48}\mathrm{C}_{2}+{}^{4}\mathrm{C}_{4}\times{}^{48}\mathrm{C}_{1}
\displaystyle =4\times194580+6\times17296+4\times1128+1\times48
\displaystyle =778320+103776+4512+48
\displaystyle =886656
\displaystyle \therefore \text{The required number of }5\text{-card combinations is }886656.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{We wish to select }6\text{ persons from }8,\text{ but if Person A is chosen,}
\displaystyle \text{then Person B must also be chosen. In how many ways can the selection be made?}
\displaystyle \text{Answer:}
\displaystyle \text{Case 1: Person A is selected.}
\displaystyle \text{Since Person B must also be selected, A and B occupy }2\text{ places.}
\displaystyle \text{The remaining }4\text{ persons are selected from the other }6\text{ persons.}
\displaystyle \text{Number of ways}={}^{6}\mathrm{C}_{4}
\displaystyle =\frac{6!}{4!\,2!}=15
\displaystyle \text{Case 2: Person A is not selected.}
\displaystyle \text{All }6\text{ persons are selected from the remaining }7\text{ persons.}
\displaystyle \text{Number of ways}={}^{7}\mathrm{C}_{6}
\displaystyle =\frac{7!}{6!\,1!}=7
\displaystyle \therefore \text{Total number of ways}=15+7=22
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{In how many ways can a team of }3\text{ boys and }3\text{ girls be selected}
\displaystyle \text{from }5\text{ boys and }4\text{ girls?}
\displaystyle \text{Answer:}
\displaystyle \text{The }3\text{ boys can be selected from }5\text{ boys in }{}^{5}\mathrm{C}_{3}\text{ ways.}
\displaystyle \text{The }3\text{ girls can be selected from }4\text{ girls in }{}^{4}\mathrm{C}_{3}\text{ ways.}
\displaystyle \therefore \text{Required number of ways}={}^{5}\mathrm{C}_{3}\times{}^{4}\mathrm{C}_{3}
\displaystyle =\frac{5!}{3!\,2!}\times\frac{4!}{3!\,1!}
\displaystyle =10\times4
\displaystyle =40
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Find the number of ways of selecting }9\text{ balls from }6\text{ red balls,}
\displaystyle 5\text{ white balls and }5\text{ blue balls, if each selection consists of }3\text{ balls}
\displaystyle \text{of each colour.}
\displaystyle \text{Answer:}
\displaystyle \text{The }3\text{ red balls can be selected from }6\text{ red balls in }{}^{6}\mathrm{C}_{3}\text{ ways.}
\displaystyle \text{The }3\text{ white balls can be selected from }5\text{ white balls in }{}^{5}\mathrm{C}_{3}\text{ ways.}
\displaystyle \text{The }3\text{ blue balls can be selected from }5\text{ blue balls in }{}^{5}\mathrm{C}_{3}\text{ ways.}
\displaystyle \therefore \text{Required number of ways}
\displaystyle ={}^{6}\mathrm{C}_{3}\times{}^{5}\mathrm{C}_{3}\times{}^{5}\mathrm{C}_{3}
\displaystyle =\frac{6!}{3!\,3!}\times\frac{5!}{3!\,2!}\times\frac{5!}{3!\,2!}
\displaystyle =20\times10\times10
\displaystyle =2000
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Determine the number of }5\text{-card combinations from a deck of }52
\displaystyle \text{cards, if there is exactly one ace in each combination.}
\displaystyle \text{Answer:}
\displaystyle \text{There are }4\text{ aces and }48\text{ non-ace cards in the deck.}
\displaystyle \text{Exactly }1\text{ ace is selected from }4\text{ aces and the remaining }4\text{ cards are}
\displaystyle \text{selected from the }48\text{ non-ace cards.}
\displaystyle \therefore \text{Required number of combinations}={}^{4}\mathrm{C}_{1}\times{}^{48}\mathrm{C}_{4}
\displaystyle =\frac{4!}{1!\,3!}\times\frac{48!}{4!\,44!}
\displaystyle =4\times\frac{48\times47\times46\times45}{4\times3\times2\times1}
\displaystyle =4\times194580
\displaystyle =778320
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{In how many ways can a cricket team of }11\text{ be selected from }17
\displaystyle \text{players, of whom only }5\text{ can bowl, if the team must include exactly }4\text{ bowlers?}
\displaystyle \text{Answer:}
\displaystyle \text{There are }5\text{ bowlers and }17-5=12\text{ non-bowlers.}
\displaystyle \text{Exactly }4\text{ bowlers are selected from }5\text{ bowlers and the remaining }7\text{ players}
\displaystyle \text{are selected from the }12\text{ non-bowlers.}
\displaystyle \therefore \text{Required number of ways}={}^{5}\mathrm{C}_{4}\times{}^{12}\mathrm{C}_{7}
\displaystyle =5\times\frac{12!}{7!\,5!}
\displaystyle =5\times\frac{12\times11\times10\times9\times8}{5\times4\times3\times2\times1}
\displaystyle =5\times792
\displaystyle =3960
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{A bag contains }5\text{ black and }6\text{ red balls. Determine the number of}
\displaystyle \text{ways in which }2\text{ black and }3\text{ red balls can be selected.}
\displaystyle \text{Answer:}
\displaystyle \text{The }2\text{ black balls can be selected from }5\text{ black balls in }{}^{5}\mathrm{C}_{2}\text{ ways.}
\displaystyle \text{The }3\text{ red balls can be selected from }6\text{ red balls in }{}^{6}\mathrm{C}_{3}\text{ ways.}
\displaystyle \therefore \text{Required number of ways}={}^{5}\mathrm{C}_{2}\times{}^{6}\mathrm{C}_{3}
\displaystyle =\frac{5!}{2!\,3!}\times\frac{6!}{3!\,3!}
\displaystyle =\frac{5\times4}{2\times1}\times\frac{6\times5\times4}{3\times2\times1}
\displaystyle =10\times20
\displaystyle =200
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{In how many ways can a student choose a programme of }5\text{ courses if }9
\displaystyle \text{courses are available and }2\text{ specific courses are compulsory for every student?}
\displaystyle \text{Answer:}
\displaystyle \text{Since }2\text{ courses are compulsory, the remaining }3\text{ courses are chosen from the}
\displaystyle \text{other }7\text{ courses.}
\displaystyle \therefore \text{Required number of ways}={}^{7}\mathrm{C}_{3}
\displaystyle =\frac{7!}{3!\,4!}
\displaystyle =\frac{7\times6\times5}{3\times2\times1}
\displaystyle =35
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{A committee of }7\text{ has to be formed from }9\text{ boys and }4\text{ girls.}
\displaystyle \text{In how many ways can this be done when the committee consists of:}
\displaystyle \text{(i) exactly }3\text{ girls; (ii) at least }3\text{ girls; (iii) at most }3\text{ girls?}
\displaystyle \text{Answer:}
\displaystyle \text{i) If the committee has exactly }3\text{ girls, it must have }4\text{ boys.}
\displaystyle \therefore \text{Required number of ways}={}^{9}\mathrm{C}_{4}\times{}^{4}\mathrm{C}_{3}
\displaystyle =\frac{9!}{4!\,5!}\times\frac{4!}{3!\,1!}
\displaystyle =\frac{9\times8\times7\times6}{4\times3\times2\times1}\times4
\displaystyle =126\times4
\displaystyle =504
\displaystyle \text{ii) If the committee has at least }3\text{ girls, it may have }3\text{ or }4\text{ girls.}
\displaystyle \therefore \text{Required number of ways}
\displaystyle ={}^{9}\mathrm{C}_{4}\times{}^{4}\mathrm{C}_{3}+{}^{9}\mathrm{C}_{3}\times{}^{4}\mathrm{C}_{4}
\displaystyle =126\times4+84\times1
\displaystyle =504+84
\displaystyle =588
\displaystyle \text{iii) If the committee has at most }3\text{ girls, it may have }0,1,2\text{ or }3\text{ girls.}
\displaystyle \therefore \text{Required number of ways}
\displaystyle ={}^{9}\mathrm{C}_{7}\times{}^{4}\mathrm{C}_{0}+{}^{9}\mathrm{C}_{6}\times{}^{4}\mathrm{C}_{1}
\displaystyle \qquad+{}^{9}\mathrm{C}_{5}\times{}^{4}\mathrm{C}_{2}+{}^{9}\mathrm{C}_{4}\times{}^{4}\mathrm{C}_{3}
\displaystyle =36\times1+84\times4+126\times6+126\times4
\displaystyle =36+336+756+504
\displaystyle =1632
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{In an examination, a question paper consists of }12\text{ questions divided}
\displaystyle \text{into Part I and Part II, containing }5\text{ and }7\text{ questions respectively. A student}
\displaystyle \text{is required to attempt }8\text{ questions, selecting at least }3\text{ from each part.}
\displaystyle \text{In how many ways can the student select the questions?}
\displaystyle \text{Answer:}
\displaystyle \text{The possible selections from Parts I and II respectively are}
\displaystyle (3,5),\quad(4,4)\quad\text{and}\quad(5,3).
\displaystyle \therefore \text{Required number of ways}
\displaystyle ={}^{5}\mathrm{C}_{3}\times{}^{7}\mathrm{C}_{5}+{}^{5}\mathrm{C}_{4}\times{}^{7}\mathrm{C}_{4}
\displaystyle \qquad+{}^{5}\mathrm{C}_{5}\times{}^{7}\mathrm{C}_{3}
\displaystyle =\frac{5!}{3!\,2!}\times\frac{7!}{5!\,2!}+\frac{5!}{4!\,1!}\times\frac{7!}{4!\,3!}
\displaystyle \qquad+\frac{5!}{5!\,0!}\times\frac{7!}{3!\,4!}
\displaystyle =10\times21+5\times35+1\times35
\displaystyle =210+175+35
\displaystyle =420
\displaystyle \therefore \text{The student can select the questions in }420\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{A parallelogram is cut by two sets of }m\text{ lines parallel to its sides.}
\displaystyle \text{Find the number of parallelograms thus formed.}
\displaystyle \text{Answer:}
\displaystyle \text{There are }m\text{ additional parallel lines in each of the two directions.}
\displaystyle \text{Including the two boundary sides, each set contains }m+2\text{ parallel lines.}
\displaystyle \text{A parallelogram is formed by choosing }2\text{ lines from each set.}
\displaystyle \therefore \text{Required number of parallelograms}
\displaystyle ={}^{m+2}\mathrm{C}_{2}\times{}^{m+2}\mathrm{C}_{2}
\displaystyle =\left({}^{m+2}\mathrm{C}_{2}\right)^2
\displaystyle =\left(\frac{(m+2)(m+1)}{2}\right)^2
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{Out of }18\text{ points in a plane, no three are collinear except }5\text{ points}
\displaystyle \text{which are collinear. How many (i) straight lines (ii) triangles can be formed}
\displaystyle \text{by joining them?}
\displaystyle \text{Answer:}
\displaystyle \text{i) The number of lines obtained by joining }18\text{ points in pairs is}
\displaystyle {}^{18}\mathrm{C}_{2}=\frac{18!}{2!\,16!}=\frac{18\times17}{2}=153
\displaystyle \text{The }5\text{ collinear points give }{}^{5}\mathrm{C}_{2}=10\text{ pairs, but these pairs}
\displaystyle \text{represent only one straight line.}
\displaystyle \therefore \text{Required number of straight lines}
\displaystyle ={}^{18}\mathrm{C}_{2}-{}^{5}\mathrm{C}_{2}+1
\displaystyle =153-10+1
\displaystyle =144
\displaystyle \text{ii) The total number of ways of choosing }3\text{ points from }18\text{ points is }{}^{18}\mathrm{C}_{3}.
\displaystyle \text{The selections formed by choosing }3\text{ of the }5\text{ collinear points do not form triangles.}
\displaystyle \therefore \text{Required number of triangles}
\displaystyle ={}^{18}\mathrm{C}_{3}-{}^{5}\mathrm{C}_{3}
\displaystyle =\frac{18!}{3!\,15!}-\frac{5!}{3!\,2!}
\displaystyle =\frac{18\times17\times16}{3\times2\times1}-\frac{5\times4}{2\times1}
\displaystyle =816-10
\displaystyle =806
\displaystyle \\

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