Question

There are 10 red balls of different shades and 9 green balls of identical shade. Then the number of ways to arrange them in a row so that two green balls are together is?
A) $\left( {10!} \right){}^{11}{P_9}$
B) $\left( {10!} \right){}^{11}{C_9}$
C) $10!$
D) $10!9!$

Answer 1

First we will arrange the red balls and then the green identical balls. So these balls can be arranged in between the red balls of different shades.

Complete step by step solution:
We are given with 10 red balls of different shades. So they simply can be arranged in $10!$ ways.
Now these balls have vacant space in between them. 9 vacant spaces in them and additional two spaces one on left side and other on right side of the red balls. Means total 11 vacant spaces.
\\_R\\_R\\_R\\_R\\_R\\_R\\_R\\_R\\_R\\_R\\_
Now we observe the spaces here, where the green balls are to be arranged.
So 11 spaces and 9 balls so the number of ways they can be arranged is ${}^{11}{C_9}$.
So the total number of ways we can arrange the balls is $\left( {10!} \right){}^{11}{C_9}$

Hence option B is correct.

Note:
This can be the question of why not option A being the same as option B. The answer is permutation follows some fixed pattern whereas combination does not follow any restriction. So option B is correct.