Find the value of 1000 C 500-1000 C 501A- 1000 C502B- 1000 C503C- 1000 C500D- 1001 C501

The correct option is D ^nC_r+^nC_r 1=^n+1C_r Here, n=1000 and r=501 ⇒^1000C_500+^1000C_501=^1001C_501 Explanation for the result: ^nC_r+^nC_r 1=n!/(n r)!× r! ...

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Standard XII

Mathematics

Higher Order Derivatives

Find the valu...

Question

Find the value of $^{1000} C_{500} +^{1000} C_{501}$

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Solution

The correct option is D

$^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
Here, n=1000 and r=501
$\Rightarrow^{1000} C_{500} +^{1000} C_{501} =^{1001} C_{501}$
Explanation for the result:
$^{n} C_{r} +^{n} C_{r - 1} = \frac{n!}{(n - r)! \times r!} + \frac{n!}{(n - (r - 1))! \times (r - 1)!}$
= $\frac{n!}{(n - r)! \times (r - 1)!} (\frac{1}{r} + \frac{1}{(n - (r - 1))})$
= $\frac{n!}{(n - r)! \times (r - 1)!} \frac{(n - r + 1 + r)}{(n - (r - 1) \times r)}$
= $\frac{(n + 1) n!}{(n - r)! \times (n - (r - 1)} (\frac{1}{r (r - 1)!})$
= $\frac{(n + 1)!}{(n - r + 1)!} (\frac{1}{r!})$
= $^{n + 1} C_{r}$

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Find the value of 1000C500+1000C501

Find the value of $^{1000} C_{500} +^{1000} C_{501}$