Question 1: From a group of \displaystyle 15 cricket players, a team of \displaystyle 11 players is to be chosen. In how many ways can this be done?

Answer:

A group of \displaystyle 15 cricket players, a team of \displaystyle 11 players is to be chosen in ways

\displaystyle = ^{15} \rm C_{11} = \frac{15!}{11! 4!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365

\displaystyle \\

Question 2: How many different boat parties of \displaystyle 8 , consisting of \displaystyle 5 boys and \displaystyle 3 girls, can be made from \displaystyle 25 boys and \displaystyle 10 girls?

Answer:

Number of different boat parties of \displaystyle 8 , consisting of \displaystyle 5 boys and \displaystyle 3 girls, that can be made from \displaystyle 25 boys and \displaystyle 10 girls

\displaystyle = ^{25} \rm C_{5} \times ^{10} \rm C_{3} = \frac{25!}{5! 20!} \times \frac{10!}{3! 7!} \\ \\ = \frac{25 \times 24 \times 23 \times 22 \times 21}{5 \times 4 \times 3 \times 2 \times 1 } \times \frac{10 \times 9 \times 8 }{3 \times 2 \times 1} = 6375600

\displaystyle \\

Question 3: In how many ways can a student choose \displaystyle 5 courses out of \displaystyle 9 courses if \displaystyle 2 courses are compulsory for every student?

Answer:

\displaystyle 2 courses are compulsory for every student.

Therefore \displaystyle 3 courses are to be chosen from \displaystyle 7 possible courses.

Hence the number of ways a student can choose \displaystyle 5 courses out of \displaystyle 9 courses if \displaystyle 2 courses are compulsory for every student

\displaystyle = ^{7} \rm C_{3} = \frac{7!}{3! 4!} = \frac{7 \times 6 \times 5 }{6} = 35

\displaystyle \\

Question 4: In how many ways can a football team of \displaystyle 11 players be selected from \displaystyle 16 players? How many of these will (i) include \displaystyle 2 particular players? (ii) exclude \displaystyle 2 particular players?

Answer:

Number of ways to select \displaystyle 11 players from \displaystyle 16 players

\displaystyle = ^{16} \rm C_{11} = \frac{16!}{11! 5!} = \frac{16 \times 15 \times 14 \times 13 \times 12 }{5 \times 4 \times 3 \times 2 \times 1 } = 4368

i) If two particular players are always included, then the number of ways

\displaystyle = ^{14} \rm C_{9} = \frac{14!}{9! 5!} = \frac{14 \times 13 \times 12 \times 11 \times 10 }{5 \times 4 \times 3 \times 2 \times 1 } = 2002

ii) If two particular players are always excluded, then the number of ways

\displaystyle = ^{14} \rm C_{11} = \frac{14!}{11! 3!} = \frac{14 \times 13 \times 12 }{ 3 \times 2 \times 1 } = 364

\displaystyle \\

Question 5: There are \displaystyle 10 professors and \displaystyle 20 students out of whom a committee of \displaystyle 2 professors and \displaystyle 3 students is to be formed. Find the number of ways in which this can be done. Further, find in how many of these committees:

(i) a particular professor is included.

(iii) a particular student is excluded.

(ii) a particular student is included.

Answer:

Number of ways committee can be formed

\displaystyle = ^{10} \rm C_{2} \times ^{20} \rm C_{3} = \frac{10!}{2! 8!} \times \frac{20!}{3! 17!} = \frac{10 \times 9}{2} \times \frac{20 \times 19 \times 18}{6} = 51300

i) Number of ways committee can be formed if a particular professor is included

\displaystyle = ^{9} \rm C_{1} \times ^{20} \rm C_{3} = \frac{9!}{1! 8!} \times \frac{20!}{3! 17!} = \frac{9}{1} \times \frac{20 \times 19 \times 18}{6} = 10260

ii) Number of ways committee can be formed if a particular student is excluded

\displaystyle = ^{10} \rm C_{2} \times ^{19} \rm C_{3} = \frac{10!}{2! 8!} \times \frac{19!}{3! 16!} = \frac{10 \times 9}{2} \times \frac{19 \times 18 \times 17}{6} = 43605

iii) Number of ways committee can be formed if a particular student is included

\displaystyle = ^{10} \rm C_{2} \times ^{19} \rm C_{2} = \frac{10!}{2! 8!} \times \frac{19!}{2! 17!} = \frac{10 \times 9}{2} \times \frac{19 \times 18 }{2} = 7695

\displaystyle \\

Question 6: How many different products can be obtained by multiplying two or more of the numbers 3, 5, 7, 11 (without repetition)?

Answer:

The number of ways products can be obtained by multiplying two or more of the number \displaystyle 3, 5, 7, 11

\displaystyle = ^{4} \rm C_{2} + ^{4} \rm C_{3} + ^{4} \rm C_{4} = 6 + 4 + 1 = 11

\displaystyle \\

Question 7: From a class of \displaystyle 12 boys and \displaystyle 10 girls, \displaystyle 10 students are to be chosen for a competition; at least including \displaystyle 4 boys and \displaystyle 4 girls. The \displaystyle 2 girls who won the prizes last year should be included. In how many ways can the selection be made?

Answer:

Given \displaystyle 2 girls who won the prizes last year should be included.

Therefore we have to select from \displaystyle 12 boys and \displaystyle 8 girls. Therefore the combinations would be

\displaystyle = ^{12} \rm C_{6} \times ^{8} \rm C_{2} + ^{12} \rm C_{5} \times ^{8} \rm C_{3} + ^{12} \rm C_{4} \times ^{8} \rm C_{4}

\displaystyle = \frac{12!}{6! 6!} \times \frac{8!}{2! 6!} + \frac{12!}{5! 7!} \times \frac{8!}{3! 5!} + \frac{12!}{4! 8!} \times \frac{8!}{4! 4!}   

\displaystyle = 924 \times 28 + 792 \times 56 + 495 \times 70

\displaystyle = 104874

\displaystyle \\

Question 8: How many different selections of 4 books can be made from 10 different books, if (i) there is no restriction; (ii) two particular books are always selected (iii) two particular books are never selected?

Answer:

i) Number of ways in which \displaystyle 4 books can be made from \displaystyle 10 different books if there is no restriction \displaystyle = ^{10} \rm C_{4} = \frac{10!}{4! 6!} = \frac{ 10 \times 9 \times 8 \times 7 }{ 4 \times 3 \times 2 \times 1 } = 210

ii) Number of ways in which \displaystyle 4 books can be made from \displaystyle 10 different books if two particular books are always selected \displaystyle = ^{8} \rm C_{2} = \frac{8!}{2! 6!} = \frac{ 8 \times 7 }{ 2 \times 1 } = 28

iii) Number of ways in which \displaystyle 4 books can be made from \displaystyle 10 different books if two particular books are never selected \displaystyle = ^{8} \rm C_{4} = \frac{8!}{4! 4!} = \frac{ 8 \times 7 \times 6 \times 5 }{ 4 \times 3 \times 2 \times 1 } = 70

\displaystyle \\

Question 9: From \displaystyle 4 officers and \displaystyle 8 soldiers in how many ways can \displaystyle 6 be chosen (i) to include exactly one officer (ii) to include at least one officer?

Answer:

Number of ways in which from \displaystyle 4 officers and \displaystyle 8 soldiers in how many ways can 6 be chosen if there is no restriction

\displaystyle = ^{12} \rm C_{6} = \frac{12!}{6! 6!} = \frac{ 12 \times 11 \times 9 \times 8 \times 7}{ 6 \times 5 \times 4 \times 3 \times 2 \times 1 } = 924

i) Number of ways in which from \displaystyle 4 officers and \displaystyle 8 soldiers in how many ways can 6 be chosen to include exactly one officer

\displaystyle = ^{4} \rm C_{1} \times ^{8} \rm C_{5} = \frac{4!}{1! 3!} \times \frac{8!}{5! 3!} = 4 \times \frac{8 \times 7 \times 6 }{3 \times 2 \times 1} = 224

ii) Number of ways in which from \displaystyle 4 officers and \displaystyle 8 soldiers in how many ways can 6 be chosen to include at least one officer

\displaystyle = ^{4} \rm C_{1} \times ^{8} \rm C_{5} + ^{4} \rm C_{2} \times ^{8} \rm C_{4}+^{4} \rm C_{3} \times ^{8} \rm C_{3}+^{4} \rm C_{4} \times ^{8} \rm C_{2}

\displaystyle = \frac{4!}{1! 3!} \times \frac{8!}{5! 3!} + \frac{4!}{2! 2!} \times \frac{8!}{4! 4!} + \frac{4!}{3! 1!} \times \frac{8!}{3! 5!} + \frac{4!}{4! 0!} \times \frac{8!}{2! 6!}   

\displaystyle = \Big( \frac{ 4 \times 8 \times 7 \times 6 }{ 3 \times 2 } \Big) + \Big( \frac{ 4 \times 3 \times 8 \times 7 \times 6 \times 5}{ 2 \times 4 \times 3 \times 2 } \Big) + \Big( \frac{ 4 \times 8 \times 7 \times 6 }{ 3 \times 2 } \Big) + \Big( \frac{ 8 \times 7 }{ 2 \times 1 } \Big)

\displaystyle = (4 \times 8 \times 7 ) + (4 \times 3 \times 7 \times 5 ) + (4 \times 8 \times 7 ) + (4 \times 7 )

\displaystyle = 224 + 420 + 224 + 28 = 896

\displaystyle \\

Question 10: A sports team of \displaystyle 11 students is to be constituted, choosing at least \displaystyle 5 from class XI and at least \displaystyle 5 from class XII. If there are \displaystyle 20 students in each of these classes, in how many ways can the teams be constituted?

Answer:

The number of ways a sports team of \displaystyle 11 students is to be constituted, choosing at least \displaystyle 5 from class XI and at least \displaystyle 5 from class XII from \displaystyle 20 students in each of these classes

\displaystyle = ^{20} \rm C_{5} \times ^{20} \rm C_{6} + ^{20} \rm C_{6} \times ^{20} \rm C_{5}

\displaystyle = \frac{20!}{5! 15!} \times \frac{20!}{6! 14!} + \frac{20!}{6! 14!} \times \frac{20!}{5! 15!}   

\displaystyle = 2 \times \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1 } \times \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1 }   

\displaystyle = 2 \times 19 \times 3 \times 17 \times 16 \times 19 \times 17 \times 8 \times 15

\displaystyle = 1201870080

\displaystyle \\

Question 11: A student has to answer \displaystyle 10 questions, choosing at least \displaystyle 4 from each of \displaystyle \text{Part A} and \displaystyle \text{Part B} . If there are \displaystyle 6 questions in \displaystyle \text{Part A} and \displaystyle 7 in \displaystyle \text{Part B} , in how many ways can the student choose \displaystyle 10 questions?

Answer:

Total number of questions \displaystyle = 10

Questions in \displaystyle \text{Part A} = 6 , Questions in \displaystyle \text{Part B} = 7

The number of ways the student choose \displaystyle 10 questions

\displaystyle = ^{6} \rm C_{4} \times ^{7} \rm C_{6} + ^{6} \rm C_{5} \times ^{7} \rm C_{5} + ^{6} \rm C_{6} \times ^{7} \rm C_{4}

\displaystyle = \frac{6!}{4! 2!} \times \frac{7!}{6! 1!} + \frac{6!}{5! 1!} \times \frac{7!}{5! 2!} + \frac{6!}{6! 0!} \times \frac{7!}{4! 3!}   

\displaystyle = \frac{6 \times 5 \times 7 }{2} + \frac{6 \times 7 \times 6 }{2} + \frac{7 \times 6 \times 5 }{3 \times 2}   

\displaystyle = 105 + 126 + 35

\displaystyle = 266

\displaystyle \\

Question 12: In an examination, a student has to answer \displaystyle 4 questions out of \displaystyle 5 questions; questions \displaystyle 1 and \displaystyle 2 are however compulsory. Determine the number of ways in which the student can make the choice.

Answer:

Total number of questions \displaystyle = 5

Number of questions to be answered \displaystyle = 4

Questions \displaystyle 1 and \displaystyle 2 are however compulsory. Therefore the student has to choose \displaystyle 2 questions from the remainder \displaystyle 3 questions.

Number of ways to choose \displaystyle 2 questions from the remainder \displaystyle 3 questions \displaystyle = ^{3} \rm C_{2} = \frac{3!}{2! 1!} = 3

\displaystyle \\

Question 13: A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. In how many ways can he choose the 7 questions?

Answer:

Total number of questions \displaystyle = 5

Number of questions to be answered \displaystyle = 4

Number of ways can he choose the \displaystyle = 7 questions

\displaystyle = ^{6} \rm C_{2} \times ^{6} \rm C_{5} + ^{6} \rm C_{3} \times ^{6} \rm C_{4} + ^{6} \rm C_{4} \times ^{6} \rm C_{3} + ^{6} \rm C_{5} \times ^{6} \rm C_{2}

\displaystyle = \frac{6!}{2! 4!} \times \frac{6!}{5! 1!} + \frac{6!}{3! 3!} \times \frac{6!}{4! 2!} + \frac{6!}{4! 2!} \times \frac{6!}{3! 3!} + \frac{6!}{5! 1!} \times \frac{6!}{2! 4!}   

\displaystyle = \frac{6 \times 5 }{2} \times \frac{6 }{1} + \frac{6 \times 5 \times 4 }{3 \times 2} \times \frac{6 \times 5 }{ 2} + \frac{6 \times 5 }{ 2} \times \frac{6 \times 5 \times 4 }{ 3 \times 2} + \frac{6 }{ 1} \times \frac{6 \times 5 }{ 2}   

\displaystyle = 105 + 126 + 35

\displaystyle = 266

\displaystyle \\

Question 14: There are \displaystyle 10 points in a plane of which \displaystyle 4 are collinear. How many different straight lines can be drawn by joining these points.

Answer:

Number of straight lines that can be drawn within \displaystyle 10 points, taking \displaystyle 2 points at a time \displaystyle = ^{10} \rm C_{2} = \frac{10!}{2! 8!} = \frac{10 \times 9}{ 2} = 45

Number of straight lines that can be drawn within \displaystyle 4 collinear points, taking \displaystyle 2 points at a time \displaystyle = ^{4} \rm C_{2} = \frac{4!}{2! 2!} = \frac{4 \times 3}{ 2} = 6

However, the \displaystyle 4 collinear points will form a straight line.

Hence the total number of lines \displaystyle = 45 - 6 + 1 = 40

\displaystyle \\

Question 15: Find the number of diagonals of (i) a hexagon (ii) a polygon of \displaystyle 16 sides.

Answer:

Note: A polygon has \displaystyle n vertices. When you join two sides, you will either get a diagonal or a side.

i) Number of diagonals of a hexagon with \displaystyle 6 sides

\displaystyle = ^{6} \rm C_{2} - 6 = \frac{6!}{2! 4!} - 6 = \frac{6 \times 5}{ 2} - 6 = 15 - 6 = 9

ii) Number of diagonals of a polygon with \displaystyle 16 sides

\displaystyle = ^{16} \rm C_{2} - 16 = \frac{16!}{2! 14!} - 16 = \frac{16 \times 15}{ 2} - 16 = 120 - 16 = 104

\displaystyle \\

Question 16: How many triangles can be obtained by joining 12 points, five of which are collinear?

Answer:

The number of triangles that can be obtained by joining \displaystyle 12 points, five of which are collinear

\displaystyle = ^{12} \rm C_{3} - ^{5} \rm C_{3} = \frac{12!}{3! 9!} - \frac{5!}{3! 2!} = \frac{12 \times 11 \times 10 }{ 3 \times 2} - \frac{5 \times 4}{ 2} = 220 - 10 = 210

\displaystyle \\

Question 17: In how many ways can a committee of \displaystyle 5 persons be formed out of \displaystyle 6 men and \displaystyle 4 women when at least one woman has to be necessarily selected ?

Answer:

The number of ways a a committee of \displaystyle 5 persons be formed out of \displaystyle 6 men and \displaystyle 4 women when at least one woman has to be necessarily selected

\displaystyle = ^{4} \rm C_{1} \times ^{6} \rm C_{4} + ^{4} \rm C_{2} \times ^{6} \rm C_{3} + ^{4} \rm C_{3} \times ^{6} \rm C_{2} + ^{4} \rm C_{4} \times ^{6} \rm C_{1}

\displaystyle = \frac{4!}{1! 3!} \times \frac{6!}{4! 2!} + \frac{4!}{2! 2!} \times \frac{6!}{3! 3!} + \frac{4!}{3! 1!} \times \frac{6!}{2! 4!} + \frac{4!}{4! 0!} \times \frac{6!}{1! 5!}   

\displaystyle = \frac{4 }{1} \times \frac{6 \times 5}{ 2 \times 1} + \frac{4 \times 3 }{ 2 \times 1} \times \frac{6 \times 5 \times 4 }{ 3 \times 2 \times 1} + \frac{4 }{1} \times \frac{6 \times 5}{ 2 \times 1} + 6

\displaystyle = 60+120+60+6 = 246

\displaystyle \\

Question 18: In a village, there are \displaystyle 87 families of which \displaystyle 52 families have at most \displaystyle 2 children. In a rural development program, \displaystyle 20 families are to be helped chosen for assistance, of which at least \displaystyle 18 families must have at most \displaystyle 2 children. In how many ways can the choice be made ?

Answer:

There are \displaystyle 52 families that have at most \displaystyle 2 children.

There are \displaystyle 35 families that have more than \displaystyle 2 children.

Selection of \displaystyle 20 families of which at least \displaystyle 18 families must have at most \displaystyle 2 children

\displaystyle = ^{52} \rm C_{18} \times ^{35} \rm C_{2} + ^{52} \rm C_{19} \times ^{35} \rm C_{1} + ^{52} \rm C_{20} \times ^{35} \rm C_{0}

\displaystyle \\

Question 19: A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girl? (ii) at least one boy and one girl? (iii) at least 3 girls?

Answer:

Number of girls \displaystyle = 4

Number of Boys \displaystyle = 7

i) Number of ways \displaystyle 5 members be selected if the team has no girl

\displaystyle = ^{7} \rm C_{5} = \frac{7!}{5! 2!} = \frac{7 \times 6 }{ 2 \times 1} = 21

ii) Number of ways \displaystyle 5 members be selected if the team has at least \displaystyle 1 boy and \displaystyle 1 girls

\displaystyle = ^{4} \rm C_{1} \times ^{7} \rm C_{4} + ^{4} \rm C_{2} \times ^{7} \rm C_{3} + ^{4} \rm C_{3} \times ^{7} \rm C_{2} + ^{4} \rm C_{4} \times ^{7} \rm C_{1}

\displaystyle = \frac{4!}{1! 3!} \times \frac{7!}{4! 3!} + \frac{4!}{2! 2!} \times \frac{7!}{3! 4!} + \frac{4!}{3! 1!} \times \frac{7!}{2! 5!} + \frac{4!}{4! 0!} \times \frac{7!}{1! 6!}   

\displaystyle = 140 + 210 + 84 + 7

\displaystyle = 441

iii) Number of ways \displaystyle 5 members be selected if the team has at least \displaystyle 3 girls

\displaystyle = ^{4} \rm C_{3} \times ^{7} \rm C_{2} + ^{4} \rm C_{4} \times ^{7} \rm C_{1}

\displaystyle = \frac{4!}{3! 1!} \times \frac{7!}{2! 5!} + \frac{4!}{4! 0!} \times \frac{7!}{1! 6!}   

\displaystyle = 84 + 7

\displaystyle = 91

\displaystyle \\

Question 20: A committee of \displaystyle 3 persons is to be constituted from a group of \displaystyle 2 men and \displaystyle 3 women. In how many ways can this be done? How many of these committees would consist of \displaystyle 1 man and \displaystyle 2 women?

Answer:

i) Number of ways committee can be selected

\displaystyle = ^{5} \rm C_{3} = \frac{5!}{3! 2!} = \frac{5 \times 4 }{ 2 \times 1} = 10

ii) Number of ways committee can be selected if consists of \displaystyle 1 man and \displaystyle 2 women

\displaystyle = ^{2} \rm C_{1} \times ^{3} \rm C_{2} = \frac{2!}{1! 1!} \times \frac{3!}{2! 1!} = 6

\displaystyle \\

Question 21: Find the number of (i) diagonals (ii) triangles formed in a decagon.

Answer:

Number of sides in a decagon \displaystyle = 10

i) Number of diagonals \displaystyle = ^{10} \rm C_{2} - 10 = \frac{10!}{2! 8!} - 10 = \frac{10 \times 9 }{ 2 \times 1} - 10 = 45 - 10 = 35

\displaystyle \\

Question 22: Determine the number of \displaystyle 5 cards combinations out of a deck of \displaystyle 52 cards if at least one of the \displaystyle 5 cards has to be a king ?

Answer:

Total number of cards in a deck \displaystyle = 52

Number of kings in a deck \displaystyle = 4

Therefore number of ways 5 cards combinations out of a deck of \displaystyle 52 cards can be drawn if at least one of the \displaystyle 5 cards has to be a king

\displaystyle = ^{48} \rm C_{1} \times ^{4} \rm C_{4} + ^{48} \rm C_{2} \times ^{4} \rm C_{3} + ^{48} \rm C_{3} \times ^{4} \rm C_{2} + ^{48} \rm C_{4} \times ^{4} \rm C_{1}

\displaystyle = \frac{48!}{1! 47!} \times \frac{4!}{4! 0!} + \frac{48!}{2! 46!} \times \frac{4!}{3! 1!} + \frac{48!}{3! 45!} \times \frac{4!}{2! 2!} + \frac{48!}{4! 44!} \times \frac{4!}{1! 3!}   

\displaystyle = 48 \times 1 + \frac{48 \times 47 }{ 2 \times 1} \times \frac{4 }{ 2 \times 1} + \frac{48 \times 47 \times 46 }{ 3 \times 2 \times 1} \times \frac{4 \times 3 }{ 2 \times 1} + \frac{48 \times 47 \times 46 \times 45 }{ 4 \times 3 \times 2 \times 1} \times \frac{4 }{ 1}   

\displaystyle = 48 + 4512 + 103776 + 778320

\displaystyle = 886656

\displaystyle \\

Question 23: We wish to select 6 persons from 8, but if the \displaystyle \text{Person A} is chosen, then \displaystyle \text{Person B} must be chosen. In how many ways can the selection be made ?

Answer:

We need to select \displaystyle 6 persons from \displaystyle 8

Number of ways persons can be selected when \displaystyle \text{Person A} is selected

\displaystyle = ^{6} \rm C_{4} = \frac{6!}{4! 2!} = 15

Number of ways persons can be selected when \displaystyle \text{Person A} is not selected

\displaystyle = ^{7} \rm C_{6} = \frac{7!}{6! 1!} = 7

Hence the total number of ways \displaystyle = 15 + 7 = 22

\displaystyle \\

Question 24: In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Answer:

Number of ways in which \displaystyle 3 boys and \displaystyle 3 girls be selected from \displaystyle 5 boys and \displaystyle 4 girls

\displaystyle = ^{5} \rm C_{3} \times ^{4} \rm C_{3} = \frac{5!}{3! 2!} \times \frac{4!}{3! 1!} = 10 \times 4 = 40

\displaystyle \\

Question 25: Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each color.

Answer:

Number of ways of selecting \displaystyle 9 balls from \displaystyle 6 red balls, \displaystyle 5 white balls and \displaystyle 5 blue balls if each selection consists of \displaystyle 3 balls of each color

\displaystyle = ^{6} \rm C_{3} \times ^{5} \rm C_{3} \times ^{5} \rm C_{3} = \frac{6!}{3! 3!} \times \frac{5!}{3! 2!} \times \frac{5!}{3! 2!} = \frac{6 \times 5 \times 4 }{ 3 \times 2 \times 1} \times \frac{5 \times 4 }{ 2 \times 1} \times \frac{5 \times 4 }{ 2 \times 1}   

\displaystyle = 20 \times 10 \times 10 = 2000

\displaystyle \\

Question 26: Determine the number of 5 cards combinations out of a deck of 52 cards if there is exactly one ace in each combination

Answer:

Total number of cards in a deck \displaystyle = 52

Number of Aces in a deck \displaystyle = 4

Therefore the number of ways in which \displaystyle 5 cards combinations out of a deck of \displaystyle 52 cards if there is exactly one ace in each combination

\displaystyle = ^{4} \rm C_{1} \times ^{48} \rm C_{4} = \frac{4!}{1! 3!} \times \frac{48!}{4! 44!} = 4 \times \frac{48 \times 47 \times 46 \times 45 }{ 4\times 3 \times 2 \times 1} = 778320

\displaystyle \\

Question 27: In how many ways can one select a cricket team of eleven from 17 players in which only 5 persons can bowl if each cricket team of 11 must include exactly 4 bowlers?

Answer:

Number of ways one can select a cricket team of eleven from \displaystyle 17 players in which only \displaystyle 5 persons can bowl if each cricket team of \displaystyle 11 must include exactly \displaystyle 4 bowlers

\displaystyle = ^{5} \rm C_{4} \times ^{12} \rm C_{7} = 5 \times \frac{12!}{7! 5!} = 5 \times \frac{12 \times 11 \times 10 \times 9 \times 8 }{5 \times 4 \times 3 \times 2 \times 1 } = 5 \times 792 = 3960

\displaystyle \\

Question 28: A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Answer:

The number of ways \displaystyle 2 black and \displaystyle 3 red balls can be selected from a bag containing \displaystyle 5 black and \displaystyle 6 red balls

\displaystyle = ^{5} \rm C_{2} \times ^{6} \rm C_{3} = \frac{5!}{2! 3!} \times \frac{6!}{3! 3!} = \frac{5 \times 4}{2 \times 1} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 200

\displaystyle \\

Question 29: In how many ways can a student choose a program of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Answer:

The number of ways can a student can choose a program of \displaystyle 5 courses if \displaystyle 9 courses are available and \displaystyle 2 specific courses are compulsory for every student

\displaystyle = ^{7} \rm C_{3} = \frac{7!}{3! 4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

\displaystyle \\

Question 30: A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consist of :

(i) exactly 3 girls? (ii) at least 3 girls? (iii) at most 3 girls?

Answer:

i) Number of ways to select the committee with exactly 3 girls

\displaystyle = ^{9} \rm C_{4} \times ^{4} \rm C_{3} = \frac{9!}{4! 5!} \times \frac{4!}{3! 1!} = \frac{9 \times 8 \times 7 \times 6 }{4 \times 3 \times 2 \times 1 } \times 4 = 504

ii) Number of ways to select the committee with at least 3 girls

\displaystyle = ^{9} \rm C_{4} \times ^{4} \rm C_{3} + ^{9} \rm C_{3} \times ^{4} \rm C_{4}

\displaystyle = \frac{9!}{4! 5!} \times \frac{4!}{3! 1!} + \frac{9!}{3! 6!} \times \frac{4!}{4! 0!}   

\displaystyle = \frac{9 \times 8 \times 7 \times 6 }{4 \times 3 \times 2 \times 1 } \times 4 + \frac{9 \times 8 \times 7 }{ 3 \times 2 \times 1 } \times 1

\displaystyle = 504+84

\displaystyle = 588

ii) Number of ways to select the committee with at most 3 girls

\displaystyle = ^{9} \rm C_{7} \times ^{4} \rm C_{0} + ^{9} \rm C_{6} \times ^{4} \rm C_{1} + ^{9} \rm C_{5} \times ^{4} \rm C_{2} + ^{9} \rm C_{4} \times ^{4} \rm C_{3}

\displaystyle = \frac{9!}{7! 2!} \times \frac{4!}{0! 4!} + \frac{9!}{6! 3!} \times \frac{4!}{1! 3!} + \frac{9!}{5! 4!} \times \frac{4!}{2! 2!} + \frac{9!}{4! 5!} \times \frac{4!}{3! 1!}   

\displaystyle = \frac{9 \times 8}{2 \times 1} \times 1 + \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times 4 + \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1} + \frac{9 \times 8 \times 7 \times 6 \times 5}{4\times 3\times 2\times 1 } \times 4

\displaystyle = 36 + 336+756+504

\displaystyle = 1632

\displaystyle \\

Question 31: In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

Answer:

Number of ways the student can select the questions

\displaystyle = ^{5} \rm C_{3} \times ^{7} \rm C_{5} + ^{5} \rm C_{4} \times ^{7} \rm C_{4} + ^{5} \rm C_{5} \times ^{7} \rm C_{3}

\displaystyle = \frac{5!}{3! 2!} \times \frac{7!}{5! 2!} + \frac{5!}{4! 1!} \times \frac{7!}{4! 3!} + \frac{5!}{5! 0!} \times \frac{7!}{3! 5!}   

\displaystyle = \frac{5 \times 4}{2\times 1} \times \frac{7 \times 6}{2 \times 1} + 5 \times \frac{7 \times 6 \times 5 }{3 \times 2 \times 1 } + 1 \times \frac{7 \times 6}{3 \times 2 \times 1}   

\displaystyle = 210 + 175 + 35

\displaystyle = 420

\displaystyle \\

Question 32: A parallelogram is cut by two sets of \displaystyle m lines parallel to its sides. Find the number of parallelograms thus formed.

Answer:

In a parallelogram, there are two sets of parallel lines.

Now, each set of parallel lines are equal to \displaystyle (m+2) lines.

Hence the number of parallelograms \displaystyle = ^{m+2} \rm C_{2} \times ^{m+2} \rm C_{2} = (^{m+2} \rm C_{2} )^2

\displaystyle \\

Question 33: Out of 18 points in a plane, no three are in the same straight line except five points which are collinear. How many (i) straight lines (ii) triangles can be formed by joining them?

Answer:

Number of triangles formed

\displaystyle = ^{18} \rm C_{3} - ^{5} \rm C_{3} = \frac{18!}{3! 15!} - \frac{5!}{3! 2!} = \frac{18 \times 17 \times 16 }{3 \times 2 \times 1 } - \frac{5 \times 4}{2 \times 1 } = 816 - 10 = 806