If `\ ^(n+1)C_3=2.\ \ ^n C_2\ t h e n=` `3` b. `4` c. `6` d. 5

Explanation

`{}^{n+1} c_{3}=2^{n} c_{2}`
`rightarrow frac{(n+1)(n)(n-1)}{3 times 2 times 1} n^{-2} c_{0}
=not frac{n(n-1)}{not 2}`
`-n_{c_{r}}
=n^{-1} c_{r-1} * frac{n}{r}-n_{c_{0}}=1`
`=frac{n(n+1)(n-1)}{3 x^{2}}
=n(n-1)`
`=n(n-1)left[frac{n+1}{6}-1right]=0`
`


=frac{n(n-1)(n-5)}{6}=0`
`=n=0,1,5`
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