A committee of 2 men and 2 women is to be formed f – Class 11 Business Mathematics | ConceptFirst

Exercises · Q10

Q. A committee of 2 men and 2 women is to be formed from 4 men and 4 women. In how many ways can this be done?

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This problem splits into two independent, unordered selections:

Choosing 2 men out of 4:

^{4}C_{2} = \frac{4!}{2!\,2!} = \frac{4\times3}{2\times1} = 6

Choosing 2 women out of 4:

$^{4}C_{2} = \frac{4!}{2!\,2!} = 6 \quad \text{(same computation, same value)}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CA Foundation 2026Set jan-20261 markMCQ
${}^{n}C_{r-1}=28$
›Reveal solutionSolution

From $56/28=2$ and $70/56=5/4$ we get $n+1=3r$ and $4n=9r+5$ , giving $r=3,\ n=8$ .

Step 1 — first ratio
$\frac{{}^nC_r}{{}^nC_{r-1}}=\frac{n-r+1}{r}=\frac{56}{28}=2\ \Rightarrow\ n-r+1=2r\ \Rightarrow\ n+1=3r.$
Step 2 — second ratio
$\frac{{}^nC_{r+1}}{{}^nC_r}=\frac{n-r}{r+1}=\frac{70}{56}=\frac54\ \Rightarrow\ 4(n-r)=5(r+1)\ \Rightarrow\ 4n=9r+5.$
Step 3 — solve

From Step 1, $n=3r-1$ . Substitute:
$4(3r-1)=9r+5\ \Rightarrow\ 12r-4=9r+5\ \Rightarrow\ 3r=9\ \Rightarrow\ r=3,\ n=8.$
Step 4 — verify

${}^8C_2=28,\quad {}^8C_3=56,\quad {}^8C_4=70.\ \checkmark$ …
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CA Foundation 2026Set may-20261 markMCQ
$Q. A group consists of 7 men and 5 women. In how many ways can a group of 4 members be selected if the group has no women? (A) 70 (B) 30 (C) 24 (D) 35$
›Reveal solutionSolution

A committee with no women is drawn entirely from the 7 men: $\binom{7}{4} = 35$ .

Step 1 — Reduce the pool

If the group of 4 can have no women, the 5 women are excluded; all 4 come from the 7 men. Selection (not arrangement), so use combinations:
$\binom{n}{r} = \frac{n!}{r!\,(n-r)!}$
Step 2 — Apply with $n=7$ , $r=4$

Use $\binom{7}{4}=\binom{7}{3}$ for an easier computation:

$\binom{7}{4} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35$ …
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CA Foundation 2026Set may-20261 markMCQ
$Q. Four cards are drawn at random from a standard deck of 52 playing cards without replacement. In how many ways it can be done such that the selected cards consist of exactly one Jack and three Aces? (A) 2304 (B) 2440 (C) 2260 (D) 2164$
›Reveal solutionSolution

Choosing exactly one Jack and three Aces = $\binom{4}{1}\binom{4}{3}=16$ . The printed options do not contain 16, so ICAI graced the question.

Setting up the count

A standard pack has 4 Jacks and 4 Aces. We must choose one of the 4 Jacks and three of the 4 Aces, independently, so we multiply the two selections:
$\binom{4}{1}\times\binom{4}{3} = 4 \times 4 = 16.$
Why this question was graced …
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CA Foundation 2025Set jan-20251 markMCQ
$Q. A panel has total of 11 members including 5 males and 6 females. Find out the number of ways of picking 2 males and 3 females from the given panel team. (A) 110 (B) 200 (C) 220 (D) 350$
›Reveal solutionSolution

$\binom{5}{2} \times \binom{6}{3} = 10 \times 20 = 200$ .

Step 1 — Choose the males
$\binom{5}{2} = \frac{5!}{2!\,3!} = 10$
Step 2 — Choose the females
$\binom{6}{3} = \frac{6!}{3!\,3!} = 20$
Step 3 — Multiply (independent choices)
$10 \times 20 = 200$
Why the other options are wrong: (A) 110 and (C) 220 come from a wrong combination value; (D) 350 over-counts, e.g. by treating a selection as ordered. …
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CA Foundation 2025Set jan-20251 markMCQ
$Q. In how many ways can an interview panel of 3 members be formed from 3 engineers, 2 psychologists and 3 managers if at least 1 engineer must be included ? (A) 30 (B) 15 (C) 46 (D) 45$
›Reveal solutionSolution

At least 1 engineer $= \binom{8}{3} - \binom{5}{3} = 56 - 10 = 46$ .

Step 1 — Count all possible panels

Total people $= 3 + 2 + 3 = 8$ ; choose any 3:
$\binom{8}{3} = 56$
Step 2 — Count panels with NO engineer

Non-engineers $= 2 + 3 = 5$ ; choose 3 from them:
$\binom{5}{3} = 10$
Step 3 — Subtract (complement rule)
$\text{At least 1 engineer} = 56 - 10 = 46$
Why the other options are wrong: (D) 45 and (A) 30 come from case-by-case slips (double counting or missing the all-3-engineer case); (B) 15 counts only one case. …
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CA Foundation 2025Set jan-20251 markMCQ
$Q. A committee of 3 members is formed from 5 women and 3 men in such a way that it consists at least 2 members who are women. In how many different ways can it be done ? (A) 40 (B) 50 (C) 60 (D) 30$
›Reveal solutionSolution

(2W,1M) $+$ (3W) $= \binom{5}{2}\binom{3}{1} + \binom{5}{3} = 30 + 10 = 40$ .

Step 1 — Break "at least 2 women" into cases

A 3-member committee with at least 2 women means either 2 women + 1 man or 3 women.

Step 2 — Case 1: 2 women and 1 man
$\binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30$
Step 3 — Case 2: 3 women
$\binom{5}{3} = 10$
Step 4 — Add the mutually exclusive cases
$30 + 10 = 40$
Why the other options are wrong: (B) 50 and (C) 60 over-count (e.g. double-counting or adding a 1-woman case); (D) 30 gives only the 2-women case and omits the all-women case. …
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CA Foundation 2025Set may-20251 markMCQ
$Q. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed ? (A) 210 (B) 1050 (C) 25200 (D) 21400$
›Reveal solutionSolution

Select then arrange: $\binom{7}{3}\binom{4}{2}\times5! = 35\times6\times120 = 25200$ .

Step 1 — Select the letters (combinations)
$\binom{7}{3}=35,\qquad \binom{4}{2}=6$
Number of ways to pick 3 consonants and 2 vowels $=35\times6=210$ .

Step 2 — Arrange each selected group (permutation)

Each chosen set has 5 distinct letters, arranged in $5!=120$ ways.

Step 3 — Multiply selection by arrangement

$210\times120 = 25200$ …
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CA Foundation 2025Set may-20251 markMCQ
$Q. A team of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done if these teams would consist of 1 man and 2 women ? (A) 10 (B) 6 (C) 16 (D) 8$
›Reveal solutionSolution

$\binom{2}{1}\times\binom{3}{2}=2\times3=6$ .

Step 1 — Break the team into required parts

The composition is fixed at 1 man + 2 women, so count each part with combinations (order within a team does not matter).
$\binom{2}{1}=2,\qquad \binom{3}{2}=3$
Step 2 — Multiply the independent choices
$2\times3 = 6$
Why the other options are wrong: (A) 10 $=\binom{5}{3}$ counts all 3-person teams ignoring the gender split; (C) 16 and (D) 8 do not match the required product. …
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CA Foundation 2025Set sep-20251 markMCQ
$Q. In a school, for a class monitor selection, there are 6 candidates, and students need to choose up to 3 monitors. A student can vote for 1 or 2 or 3 candidates. In how many ways a student can vote ? (A) 41 (B) 42 (C) 43 (D) 44$
›Reveal solutionSolution

Total valid votes $= {}^6C_1 + {}^6C_2 + {}^6C_3 = 6 + 15 + 20 = 41$ .

Step 1 — Model each vote as a selection

Choosing candidates (order irrelevant) is a combination. A student may pick 1, 2, OR 3 of the 6 candidates, so we sum the three cases.
${}^nC_r = \frac{n!}{r!\,(n-r)!}$
Step 2 — Compute each case

Vote for Combinations Value
1 candidate ${}^6C_1$ 6
2 candidates ${}^6C_2$ 15
3 candidates ${}^6C_3$ 20

Step 3 — Add
$6 + 15 + 20 = 41$
Why the other options are wrong: (B) 42, (C) 43, (D) 44 all over-count — e.g. adding a phantom "vote for 0" or an extra selection case. …
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CA Foundation 2024Set sep-20241 markMCQ
$Q. How many total combinations can be formed of 8 different counters marked as 1, 2, 3, 4, 5, 6, 7 & 8, taking 4 counters at a time and there being at least one odd and one even numbered counter in each combination ? (A) 68 (B) 66 (C) 64 (D) 62$
›Reveal solutionSolution

Total combinations − (all odd) − (all even) = 70 − 1 − 1 = 68.

Step 1 — Count all 4-counter selections

There are 4 odd (1,3,5,7) and 4 even (2,4,6,8) counters.
$^8C_4 = \dfrac{8!}{4!\,4!} = 70$
Step 2 — Remove the selections that violate the condition

A selection is invalid if it has no even counter (all four odd) or no odd counter (all four even):
$^4C_4 = 1 \ (\text{all odd}),\qquad ^4C_4 = 1 \ (\text{all even})$
Step 3 — Subtract

$70 - 1 - 1 = 68$ …
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CA Foundation 2024Set sep-20241 markMCQ
$Q. In a party every person shakes hands with every other person. If there are 105 handshakes in total, find the number of persons in the party. (A) 14 (B) 15 (C) 21 (D) 22$
›Reveal solutionSolution

Handshakes = ⁿC₂ = n(n−1)/2 = 105 ⇒ n(n−1) = 210 ⇒ n = 15.

Step 1 — Model handshakes as pairs

A handshake is an unordered pair of distinct people:
$^nC_2 = \dfrac{n(n-1)}{2} = 105$
Step 2 — Solve the equation
$n(n-1) = 210$
Look for two consecutive integers whose product is 210: $15 \times 14 = 210$ .
$n = 15$
Why the other options are wrong: (A) 14 gives 91 handshakes; (C) 21 gives 210; (D) 22 gives 231 — none equals 105. Only n = 15 works. …
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CA Foundation 2024Set sep-20241 markMCQ
$Q. A selection is to be made for one post of Principal and two posts of Vice-Principal. Amongst the six candidates called for the interview, only two are eligible for the post of Principal, while they all six are eligible for the post of Vice-Principal. The number of possible combinations for the selection is : (A) 4 (B) 12 (C) 18 (D) 20$
›Reveal solutionSolution

Principal (2 eligible) × Vice-Principals from the rest: 2 × ⁵C₂ = 2 × 10 = 20.

Step 1 — Select the Principal

Only 2 of the 6 candidates are eligible for Principal:
$2 \text{ ways}$
Step 2 — Select the two Vice-Principals

All six are eligible for VP, but one is now the Principal, leaving 5 candidates for the 2 identical VP posts:
$^5C_2 = \dfrac{5 \times 4}{2} = 10$
Step 3 — Multiply the independent choices

$2 \times 10 = 20$ …
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