Question

There are $n$ identical red balls $\&m$ identical green balls. The number of different linear arrangements consisting of " $n$ red balls but not necessarily all the green balls" is ${}^x{C}_y$ then -

• A. $x=m+n,y=m$
• B. $x=m+n+1,y=m$
• C. $x=m+n+1,y=m+1$
• D. $x=m+n,y=n$

Answer 1

Answer: (B)

Case 1 : When all $n$ red balls are taken but no green ball.
Only 1 arrangement is possible.
Case $2:n$ red balls and 1 green balls


${}^{n+1}{C}_0+{}^{n+1}{C}_1+{}^{n+2}{C}_2+\dots \dots \dots \dots +{}^{n+m}{C}_m$
${}^{n+2}{C}_1+{}^{n+2}{C}_2+\dots \dots \dots +{}^{n+m}{C}_m$
$\left(\therefore {}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r\right)$
${}^{n+3}{C}_2+{}^{n+3}{C}_3+\dots \dots \dots +{}^{n+m}{C}_m$
Finally we get the sum as: ${}^{m+n+1}{C}_m$