The value of nCn-rnCr is

We will solve each term separately. We know that ^nC_r=n!(n r)!× r! In case of nbsp; ^nC_n r value of r changes to (n r). ⇒^nC_n r=n![n (n r)]!×(n r)! ⇒^nC_ ...

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Standard VIII

Mathematics

Notations of Combinations

The value of ...

Question

The value of $\frac{^{n} C_{n - r}}{^{n} C_{r}}$ is

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Solution

We will solve each term separately.
We know that $^{n} C_{r} = \frac{n!}{(n - r)! \times r!}$

In case of $^{n} C_{n - r}$ value of $r$ changes to $(n - r)$ .
$\Rightarrow^{n} C_{n - r} = \frac{n!}{[n - (n - r)]! \times (n - r)!}$

$\Rightarrow^{n} C_{n - r} = \frac{n!}{[n - n + r]! \times (n - r)!}$

$\Rightarrow^{n} C_{n - r} = \frac{n!}{r! \times (n - r)!} =^{n} C_{r}$

$\Rightarrow^{n} C_{n - r} =^{n} C_{r}$

$\Rightarrow \frac{^{n} C_{n - r}}{^{n} C_{r}} = 1.$

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The value of nCn−rnCr is

We will solve each term separately. We know that nCr=n!(n−r)!×r! In case of nCn−r value of r changes to (n−r). ⇒nCn−r=n![n−(n−r)]!×(n−r)! ⇒nCn−r=n![n−n+r]!×(n−r)! ⇒nCn−r=n!r!×(n−r)!=nCr ⇒nCn−r=nCr ⇒nCn−rnCr=1. ​​​​​​​​​​​​​​

The value of $\frac{^{n} C_{n - r}}{^{n} C_{r}}$ is