If n - m C 2- then the value of n C 2 isA- 2 m -1 C 2B- 3 m -1 C 4C- m -1 C 4D- 3 m -1 C 2

The correct option is B 3 ^m+1C_4We know that, n = ^mC_2 = m(m 1)2 Now, ^nC_2 = n(n 1)2 ⇒ ^nC_2 = m(m 1)(m(m 1)2 1)2× 2 ⇒ ^nC_2 = m(m 1)(m^2 m 2)8 ⇒ nbsp; ^ ...

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Standard XII

Mathematics

Pascal Triangle

If n = m C 2,...

Question

If $n =^{m} C_{2}$ , then the value of $^{n} C_{2}$ is

^{m + 1} C_{4}

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2^{m + 1} C_{2}

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3^{m + 1} C_{4}

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3^{m + 1} C_{2}

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Solution

The correct option is C $3^{m + 1} C_{4}$
We know that,
$n =^{m} C_{2} = \frac{m (m - 1)}{2}$
Now,
$^{n} C_{2} = \frac{n (n - 1)}{2} \Rightarrow^{n} C_{2} = \frac{m (m - 1) (\frac{m (m - 1)}{2} - 1)}{2 \times 2} \Rightarrow^{n} C_{2} = \frac{m (m - 1) (m^{2} - m - 2)}{8} \Rightarrow^{n} C_{2} = \frac{m (m - 1) (m + 1) (m - 2)}{8} \Rightarrow^{n} C_{2} = 4! \times \frac{(m + 1) m (m - 1) (m - 2) (m - 3)!}{8 \times 4! (m - 3)!} \Rightarrow^{n} C_{2} = 4! \times \frac{^{m + 1} C_{4}}{8} ∴^{n} C_{2} = 3^{m + 1} C_{4}$

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Pascal Triangle

Standard XII Mathematics

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If n= mC2, then the value of nC2 is

The correct option is C 3 m+1C4We know that, n= mC2=m(m−1)2 Now, nC2=n(n−1)2⇒ nC2=m(m−1)(m(m−1)2−1)2×2⇒ nC2=m(m−1)(m2−m−2)8⇒ nC2=m(m−1)(m+1)(m−2)8⇒ nC2=4!×(m+1)m(m−1)(m−2)(m−3)!8×4!(m−3)!⇒ nC2=4!× m+1C48∴ nC2=3 m+1C4

If $n =^{m} C_{2}$ , then the value of $^{n} C_{2}$ is