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 the other 2 are also together but separate from the first two.
(a) 43200
(b) 42200
(c) 84200
(d) 64400

Answer 1

Hint:Calculate the number of ways of arranging 5 boys. Calculate the number of choosing 2 spaces out of 6 spaces between boys. Calculate the number of ways of choosing 2 girls to be seated together. Calculate the number of ways of arranging the 2 pairs of girls. Multiply all these values to calculate the number of ways of arranging all the people.

Complete step-by-step answer:
We have to arrange the seating of 5 boys and 4 girls such that two girls are together and the other 2 are also together but separate from the first two.
We will first calculate the number of ways of arranging 5 boys.
We know that we can arrange ‘n’ objects in $n!$ ways.
Substituting $n=5$ in the above expression, the number of ways of arranging 5 boys is $=5!=5\times 4\times 3\times 2\times 1=120$.
We know that there is a space between any two consecutive boys. So there are 4 paces between boys and one space at each end of the line in which the boys are sitting.
Thus, the total number of spaces around boys is $=2+4=6$.
We will now choose any two spaces to seat the 2 pairs of girls.
We know that we can choose ‘r’ objects out of ‘n’ objects in ${}^{n}{{C}_{r}}$ ways.
So, the number of ways of choosing two spaces out of 6 spaces is $={}^{6}{{C}_{2}}=\dfrac{6!}{2!4!}=\dfrac{6\times 5}{2}=15$.
We will now choose any two girls out of 4 girls who can be seated together.
So, the number of ways of choosing 2 girls out of 4 girls is $={}^{4}{{C}_{2}}=\dfrac{4!}{2!2!}=\dfrac{4\times 3}{2}=6$. The remaining two girls can be seated together.
We will now calculate the number of ways to arrange the seating of 2 pairs of girls.
We can arrange each pair of girls in $=2!=2$ ways. So, we can arrange both pairs of girls in $=2\times 2=4$ ways.
We will now calculate the number of ways to seat all of them in a straight line. To do so, we will multiply all the above-calculated values.
So, we have $=120\times 15\times 6\times 4=43200$.
Hence, the number of ways to arrange the seating of 5 boys and 4 girls such that two girls are together and other 2 are also together but separate from the first two is 43200, which is option (a).

Note: We must know the combinatorial formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We can’t solve this question without using the fact that we can arrange ‘n’ objects in $n!$ ways and we can choose ‘r’ objects out of ‘n’ objects in ${}^{n}{{C}_{r}}$ ways.