CUET UG 2025 — General Test Quantitative Reasoning

The question asks: How many different handshakes can happen between 10 people?

When Person A shakes hands with Person B, it's the same handshake as when Person B shakes hands with Person A. Since order doesn't matter, combinations are used.

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The number of ways to select 2 persons from 10 persons is given by $C(10, 2)$C(10,2) or $^{10}C_2$10C2​.

The combination formula is:

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

Where $n = 10$n=10 (total number of people) and $r = 2$r=2 (people selected for handshake).

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$$C(10, 2) = \dfrac{10!}{2! \times (10-2)!}$$C(10,2)=2!×(10−2)!10!​

$$C(10, 2) = \dfrac{10!}{2! \times 8!}$$C(10,2)=2!×8!10!​

$$C(10, 2) = \dfrac{10 \times 9}{2 \times 1}$$C(10,2)=2×110×9​

$$C(10, 2) = \dfrac{90}{2}$$C(10,2)=290​

$$C(10, 2) = 45$$C(10,2)=45

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Alternatively, the first person can shake hands with 9 others, the second person can shake hands with 8 remaining others, the third person with 7 remaining others, and so on.

Total $= 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45$=9+8+7+6+5+4+3+2+1=45

Therefore, the answer is 45.