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UPSC NDA I 2026 Mathematics Paper

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Question : 10 of 120

Marks: +1, -0

If

n={ }^{m} C_{2}

{ }^{n} C_{2}

equal to ?

${ }^{m+1} C_{4}$
$2 \times{ }^{m+1} C_{4}$
$3 \times{ }^{m+1} C_{4}$
${ }^{m+2} C_{4}$

Solution:

Concept:

Use

{}^{n}C_{2} = \frac{n(n-1)}{2}

after substituting

n = {}^{m}C_{2}

.

Explanation:

Given

n = {}^{m}C_{2} = \frac{m(m-1)}{2}

.

Then,

{}^{n}C_{2} = \frac{n(n-1)}{2}

Substitute

n = \frac{m(m-1)}{2}

:

{}^{n}C_{2} = \frac{1}{2}\left(\frac{m(m-1)}{2}\right)\left(\frac{m(m-1)}{2} - 1\right)

= \frac{m(m-1)(m(m-1)-2)}{8}

= \frac{m(m-1)(m-2)(m+1)}{8}

Now,

3 \times {}^{m+1}C_{4} = 3 \times \frac{(m+1)m(m-1)(m-2)}{24}

= \frac{m(m-1)(m-2)(m+1)}{8}

Thus it matches.

Answer:

{}^{n}C_{2} = 3 \times {}^{m+1}C_{4}

Correct option: C

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