If $^{10} C_{r} =^{10} C_{r + 2}$ then $^{5} C_{r}$ equals

120

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10

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360

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5

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Solution

The correct option is D $5$
We know that $^{n} C_{x} =^{n} C_{y} \Rightarrow x + y = n$
$^{10} C_{r} =^{10} C_{r + 2} \Rightarrow r + r + 2 = 10$
$\Rightarrow 2 r = 10 - 2 = 8$
$\Rightarrow r = \frac{8}{2} = 4$
$∴^{5} C_{r} =^{5} C_{4} = \frac{5!}{1! 4!} = \frac{5 \times 4!}{4!} = 5$

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Similar questions

r

{^{10} C}_{r + 2}

{^{10} C}_{r} = 10 : 21

^{10} C_{r} = 120

n > 2

n \sum r = 2 {(- 2)}^{r} ∣ ∣ ∣ ∣ \begin{matrix} ^{n - 2} C_{r - 2} & ^{n - 2} C_{r - 1} & ^{n - 2} C_{r} - 3 & 1 & 1 2 & - 1 & 0 \end{matrix} ∣ ∣ ∣ ∣

A = [\begin{matrix} 10 & 0 4 & 10 \end{matrix}]

| A |

The number of moles of Cr and O in a compound are 4.8 x 10¹⁰ and 9.6 x 10¹⁰respectively. Its empirical formula is:

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