If $^{16} C_{2 r} =^{16} C_{r + 2^{'}}$ find $^{r} C_{4}$

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Solution

If $^{n} C_{r} =^{n} C_{p}$

then r + p = n

$∴ 16 = r + r + 2$

$r = 7$

then $^{r} C_{4} =^{7} C_{4}$ $(∵ r = 7)$

$\Rightarrow \frac{7!}{4! (7 - 4)!} (∵^{n} C_{r} = \frac{n!}{r! (n - r)!})$

$\Rightarrow \frac{7 \times 5 \times 6}{3 \times 2} = 35$

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Similar questions

^{16} C_{r} =

^{16} C_{r + 2}

^{r} C_{4}

16 \sum r = 2^{16} C_{r} =

^{34} C_{5} + 4 \sum r = 0^{(38 - r)} C_{4}

^{34} C_{5} + 4 \sum r = 0^{38 - r} C_{4}

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