If n and r are two positive integers such that n ≥ r, then n Cr-1+ n Cr is equal to:

Question

If n and r are two positive integers such that n ≥ r, then n Cr-1+ n Cr is equal to: (A)  n Cr-1 (B)  n Cr (C)  n-1 Cr (D)  n+1 Cr

Tardigrade Admin · Accepted Answer

If n ≥ r, then n Cr-1+ n Cr= n+1 Cr Now, n Cr-1+ n Cr =(n !/(n-r+1) !(r-1) !)+(n !/(n-r) ! r !) =n ![(r/(n-r+1) ! r !)+(n-r+1/(n-r+1) ! r !)] =n ![(n+1/(n-r+1) ! r !)] = n+1 Cr

Q. If $n$ and $r$ are two positive integers such that $n \geq r$ , then $^{n} C_{r - 1} +^{n} C_{r}$ is equal to:

$^{n} C_{r - 1}$

$^{n} C_{r}$

$^{n - 1} C_{r}$

$^{n + 1} C_{r}$

Q. If n and r are two positive integers such that n≥r, then nCr−1​+nCr​ is equal to:

nCr−1​

nCr​

n−1Cr​

n+1Cr​

If n≥r, then nCr−1​+nCr​=n+1Cr​ Now, nCr−1​+nCr​ =(n−r+1)!(r−1)!n!​+(n−r)!r!n!​ =n![(n−r+1)!r!r​+(n−r+1)!r!n−r+1​] =n![(n−r+1)!r!n+1​] =n+1Cr​

Q. If $n$ and $r$ are two positive integers such that $n \geq r$ , then $^{n} C_{r - 1} +^{n} C_{r}$ is equal to:

$^{n} C_{r - 1}$

$^{n} C_{r}$

$^{n - 1} C_{r}$

$^{n + 1} C_{r}$