If `\ ^(16)C_r=\ \ ^(16)C_(r+2)`, find `'^r C_4`

Explanation

Given:

`\ ^{16} C_{r}=\ ^{16} C_{r+2}` `16=r+r+2[because.` Property `5:\ ^{n} C_{x}=\ ^{n} C_{y} Rightarrow x=y` or `x+y=n`

`Rightarrow 2 r+2=16`

`Rightarrow 2 r=14`

`Rightarrow r=7` Now, `\ ^{r} C_{4}=\ ^{7} C_{4}`

`Rightarrow\ ^{7} C_{4}=\ ^{7} C_{3}[because\ ^{n} C_{r}=\ ^{n} C_{n-r}.`

`Rightarrow\ ^{7} C_{4}=\ ^{7} C_{3}=frac{7}{3} times frac{6}{2} times frac{5}{1} times\ ^{4} C_{0}`
`{[because^{n} C_{r}=frac{n}{r} cdot\ ^{n-1} C_{r-1}] }`

`Rightarrow\ ^{7} C_{4}=35[because\ ^{n} C_{0}=1]`