JEE Main 2022 — Permutation Combination Question with Solution

Given, $4$ blue & $5$ red spherical ball

Now according to question there should be minimum 2 blue ball in between any two red ball.

So assume $a,b,c,d,e,f$ are number of blue balls,

So, $a+b+c+d+e+f=11$ ......(i)

Where $b,c,d,e⩾2$

Now let $b=b^{\text{'}}+2$

$c=c^{\text{'}}+2$

$d=d^{\text{'}}+2$

$e=e^{\text{'}}+2$

So equation (i) becomes

$a+b^{\text{'}}+2+c^{\text{'}}+2+d^{\text{'}}+2+e^{\text{'}}+2+f=11$

$\Rightarrow a+b^{\text{'}}+c^{\text{'}}+d^{\text{'}}+e^{\text{'}}+f=3$

Now total ways will be

${=}^{ 3+6-1}{C}_{6-1}{=}^8{C}_5{=}^8{C}_3=56$