Henry Ernest Dudeney/Puzzles and Curious Problems/355 - The Seven Children/Solution
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Puzzles and Curious Problems by Henry Ernest Dudeney: $355$
- The Seven Children
- 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?
Solution
$1$ in $7$.
Proof
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 $3$.
From Number of Permutations again, this is:
- ${}^2 P_3 = \dfrac {3!} {\paren {3 - 2}!} = 3! = 6$
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 $7$.
$\blacksquare$
Sources
- 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $355$. -- The Seven Children
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $470$. The Seven Children