Problem 742

See problems $\underline{60342}, \underline{35628}$.

## Solution

Line up all the people and give them pair numbers: give a pair of ones to some two, a pair of twos to some two, a pair of threes to some two, ..., and a pair of sevens to some two. The number of ways to distribute $\$ 14 \$$ to the 14 people standing in a row, giving $\$ 7$ pairs such numbers, is $\$ 14$ ! : (2!)^ 7 . In this case, the methods that differ by the permutation of pairs as a whole have been counted $\$ 7!$ \$ times. Since the order in which the pairs stand does not matter, we divide:

$\$ \$$ frac $\{14!\}\{(2!) \wedge 7 \backslash c d o t ~ 7!\}=\backslash f r a c\{14$ $\cdot 13 \cdot$ ··· $\backslash$ cdot 2 $\cdot$ 1$\}\{(2$ $\cdot$ 7$)$ $\cdot$ (2 $\cdot$ 6 ) $\cdot$ (2 $\cdot$

!

## Answer

$13 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3=13!$ ! ways.