CUET UG 2025 — General Test Quantitative Reasoning

The formula for permutation is $^nP_r = \frac{n!}{(n-r)!}$nPr​=(n−r)!n!​

$$^8P_3 = \frac{8!}{(8-3)!}$$8P3​=(8−3)!8!​

$$= \frac{8!}{5!}$$=5!8!​

$$= 8 \times 7 \times 6$$=8×7×6

$$= 336$$=336

The formula for combination is $^nC_r = \frac{n!}{r!(n-r)!}$nCr​=r!(n−r)!n!​

$$^{10}C_3 = \frac{10!}{3! \times 7!}$$10C3​=3!×7!10!​

$$= \frac{10 \times 9 \times 8}{3 \times 2 \times 1}$$=3×2×110×9×8​

$$= \frac{720}{6}$$=6720​

$$= 120$$=120

Therefore:

$$^8P_3 - ^{10}C_3 = 336 - 120$$8P3​−10C3​=336−120

$$= 216$$=216

(A) matches with (III) 216

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$$^8P_5 = \frac{8!}{(8-5)!}$$8P5​=(8−5)!8!​

$$= \frac{8!}{3!}$$=3!8!​

$$= 8 \times 7 \times 6 \times 5 \times 4$$=8×7×6×5×4

$$= 6720$$=6720

(B) matches with (IV) 6720

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$$^nP_4 = n(n-1)(n-2)(n-3) = 360$$nP4​=n(n−1)(n−2)(n−3)=360

For $n = 6$n=6:

$$6 \times 5 \times 4 \times 3 = 360$$6×5×4×3=360

(C) matches with (I) 6

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$$^nC_2 = \frac{n(n-1)}{2} = 210$$nC2​=2n(n−1)​=210

$$n(n-1) = 420$$n(n−1)=420

For $n = 21$n=21:

$$21 \times 20 = 420$$21×20=420

$$\frac{420}{2} = 210$$2420​=210

(D) matches with (II) 21

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Final Matching:

(A) → (III) 216

(B) → (IV) 6720

(C) → (I) 6

(D) → (II) 21