Question

$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if two girls are together and other $2$ are also together but separate from the first two

• A. $43200$
• B. $42200$
• C. $84200$
• D. $64400$

Answer 1

The correct option is A $43200$

Number of ways in which $5$ boys can be arranged $={}^5{P}_5=120$

Now girls can be arranged as $GBGBGBGBGBG$ that is girls can occupy $6$ places as shown.

Number of ways of arranging $2$ groups of girls in $6$ places $={}^6{P}_2=30$

Now number of ways for selecting two girls from $4$ girls to form a group and then selecting the left two girls to form another group ${=}^4{C}_2{\times }^2{C}_2=6$

Also number of ways for choosing first group of girls and then second group of girls is given by ${}^2{C}_1{\times }^1{C}_1=2$

Hence the answer is $=120\times 30\times 6\times 2=43200$