Henry Ernest Dudeney/Puzzles and Curious Problems/355 - The Seven Children/Solution

Puzzles and Curious Problems by Henry Ernest Dudeney: $355$

Four boys and three girls are seated in a row at random.

What are the chances that the two children at the end of each row will be girls?

$1$ in $7$.

From Number of Permutations, there are $7! = 5040$ ways in total of seating the $7$ children.

Now, we are given that there are $3$ girls.

The number of ways of seating the two girls at either end is the number of permutations of $2$ out of $3$.

${}^2 P_3 = \dfrac {3!} {\paren {3 - 2}!} = 3! = 6$

Having seated these two girls, there are $5$ children remaining to be seated in the remaining $5$ seats.

From Number of Permutations, there are $5! = 120$ ways of doing that.

Hence there are $6 \times 120 = 720$ ways of seating the children with a girl at either end.

Therefore the chances are $720$ in $5040$, or $1$ in $7$.

$\blacksquare$