how many ways can 8 boys be divided into two unequal sets?
the answer said 92. But i dont get it!!!
Thanks
Hi differentiate,
To divide 8 boys into two unequal sets,
they may be divided into
(a) groups of 1 and 7,
(b) groups of 2 and 6,
(c) groups of 3 and 5
but not groups of 4 and 4 as they would be equal sets
and groups of 0 and 8 is one set.
There are $\displaystyle \binom{8}{1}=8_{C_1}$ ways to divide into groups of 1 and 7,
since that's the number of ways to choose 1 of the 8,
(leaving 7 in the other group).
There are $\displaystyle \binom{8}{2}=8_{C_2}$ ways to divide into groups of 2 and 6,
since that's the number of ways to choose 2 of the 8,
(leaving 6 in the other group).
There are $\displaystyle \binom{8}{3}=8_{C_3}$ ways to divide into groups of 3 and 5,
since that's the number of ways to choose 3 of the 8,
(leaving 5 in the other group).
Remember
$\displaystyle \binom{8}{1}=\binom{8}{7}$
$\displaystyle \binom{8}{2}=\binom{8}{6}$
$\displaystyle \binom{8}{3}=\binom{8}{5}$
(the number of ways to choose 1 from 8) = (the number of ways to choose 7 from 8)
(the number of ways to choose 2 from 8) = (the number of ways to choose 6 from 8)
(the number of ways to choose 3 from 8) = (the number of ways to choose 5 from 8)