If ⁿC₁₀ = ⁿC₁₂, find ²³C_n.

Open in App

Solution

Given: ⁿC₁₀ = ⁿC₁₂
We have,
$^{n} C_{10} =^{n} C_{12}$
$\Rightarrow n = 12 + 10 = 22$ [∵ $^{n} C_{x} =^{n} C_{y} \Rightarrow x = y$ or, $n = x + y$ ]

Now, $^{23} C_{22} =^{23} C_{1}$ [∵ $^{n} C_{r} =^{n} C_{n - r}$ ]
$\Rightarrow^{23} C_{22} =^{23} C_{1} = \frac{23}{1} \times^{22} C_{0}$ [∵ $^{n} C_{r} = \frac{n}{r}^{n - 1} C_{r - 1}$ ]
$\Rightarrow^{23} C_{22} = 23$ [∵ $^{n} C_{0} = 1$ ]

Suggest Corrections

Similar questions

If $^{n} C_{10} =^{n} C_{12}$ , find $^{23} C_{n}$

^{n} C_{10} =^{n} C_{12}

^{n} C_{10} =^{n} C_{15}

^{27} C_{n}

^{n} C_{8} =^{n} C_{12}

n

^{'} n^{'}

n C_{13} = n C_{12}

Join BYJU'S Learning Program