CAT 2025 Lesson : Permutations & Combinations - Forming Groups

In how many ways can a team of $5$ members be selected from $6$ women and $5$ men such that
(I) exactly $3$ of the members are women?
(II) at least $3$ of the members are women?

Solution

Case I: In the total of $5$ members, if $3$ are women, the remaining $2$ are men.
Number of ways of selecting $3$ women and $2$ men $=$ $^{6}C_{3} \times$ $^{5}C_{2}$ $= 20 \times 10 = 200$

Case II: If at least $3$ women, then we should include the cases of $4$ women & $1$ man and $5$ women & no man.

Number of ways of selecting $4$ women and $1$ man $=$ $^{5}C_{4} \times$ $^{5}C_{1}$ $= 15 \times 5 = 75$
Number of ways of selecting $5$ women and $0$ man $=$ $^{5}C_{5} \times$ $^{5}C_{0} = 6 \times 1 = 6$

Total number of selections $= 200 + 75 + 6 = 281$

Answer: (I) $200$; (II) $281$

A team of $5$ needs to be selected from $5$ Indians and $6$ Chinese. How many such selections would have at least $1$ Indian and $2$ Chinese?

Solution

Number of ways of selecting
$1$ Indian and $4$ Chinese $=$ $^{5}C_{1} \times$ $^{6}C_{4} = 5 \times 15 = 75$
$2$ Indians and $3$ Chinese $=$ $^{5}C_{2} \times$ $^{6}C_{3} = 10 \times 20 = 200$
$3$ Indians and $2$ Chinese $=$ $^{5}C_{3} \times$ $^{6}C_{2} = 10 \times 15 = 150$
Total selections $= 75 + 200 + 150 = 425$

Answer: $425$

In how many ways can you divide $10$ software programmers into $3$ groups comprising of $2, 3$ and $5$ members?

Solution

Applying the formula provided above, we permute the number of software programmers and divide by the permutations possible in each group (as this pertains to combinations or selections in a group).

$\dfrac{10!}{2! 3! 5!} = \dfrac{10 \times 9 \times 8 \times 7 \times 6}{2 \times 6} = 10 \times 9 \times 4 \times 7 = 2520$

Answer: $2520$

In how many ways can you divide $10$ software programmers into $3$ groups
(I) comprising of $3$, $3$ and $4$ members?
(II) comprising of $3, 3$ and $4$ members, that are labelled A, B and C respectively?

Solution

Case I: Here the two $3$-member groups are similar. Therefore, we need to apply the formula and divide by another $2!$, to avoid double counting.

Note: Where Group $1$ and Group $2$ are $3$-member groups, note that it does not matter whether (a, b, c) and (d, e, f) are in group $1$ and group $2$ or in group $2$ and group $1$ respectively. This is to be counted only once.

Possible divisions $= \dfrac{10!}{3! 3! 4!} \times \dfrac{1}{2!} =$ $\dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{6 \times 6 \times 2} =$ $10 \times 3 \times 2 \times 7 \times 5= 2100$

Case II: Here, all the groups are distinctly labelled A, B and C. As there is a difference between Group A and Group B, we do not divide by $2!$.

Possible divisions $= \dfrac{10!}{3! 3! 4!} =$ $\dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{6 \times 6} =$ $10 \times 3 \times 4 \times 7 \times 5= 4200$

Answer: (I) $2100$; (II) $4200$

In how many ways can you divide $8$ software programmers into $4$ groups comprising of $2$ members each?

Solution

Here we have $4$ similar groups. Therefore, we divide by another $4!$.

Possible divisions $= \dfrac{8!}{2! 2! 2! 2!} \times \dfrac{1}{4!} =$ $\dfrac{8 \times 7 \times 6 \times 5}{2 \times 2 \times 2 \times 2} =$ $7 \times 3 \times 5= 105$

Answer: $105$