CUET UG 2025 — General Test Quantitative Reasoning

The problem requires choosing 3 people out of 7 people for a committee.

In a committee, order doesn't matter. Choosing Ram, Sita, and Geeta is the same as choosing Geeta, Ram, and Sita. This means combinations are used.

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When choosing $r$r things from $n$n things where order doesn't matter:

$$C(n,r) = \dfrac{n!}{r! \times (n-r)!}$$C(n,r)=r!×(n−r)!n!​

where $n!$n! (n factorial) $= n \times (n-1) \times (n-2) \times ... \times 1$=n×(n−1)×(n−2)×...×1

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For this problem:

• $n = 7$n=7 (total people)
• $r = 3$r=3 (people to choose)

$$C(7,3) = \dfrac{7!}{3! \times 4!}$$C(7,3)=3!×4!7!​

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Instead of calculating $7! = 5040$7!=5040, the expression can be simplified:

$$C(7,3) = \dfrac{7 \times 6 \times 5 \times 4!}{3! \times 4!}$$C(7,3)=3!×4!7×6×5×4!​

The $4!$4! cancels from numerator and denominator:

$$= \dfrac{7 \times 6 \times 5}{3!}$$=3!7×6×5​

$$= \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1}$$=3×2×17×6×5​

$$= \dfrac{210}{6}$$=6210​

$$= 35$$=35

Therefore, the committee can be chosen in 35 ways.