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?

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?

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?

$\displaystyle 2$ 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?

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.

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)?

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?

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?

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?

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?

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?

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.

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?

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.

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.

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?

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 ?

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 ?

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?

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?

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.

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 ?

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 ?

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?

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.

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

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?

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.

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?

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?

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?

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.

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?

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