Dilemma for shoes

"he asked his dad to choose 5 pairs (from 8 pairs) of shoes at random." so, we get

$=8C5$

$= \frac {8!}{(5!) (8-5)!}$

$= \frac {8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(5 \times 4 \times 3 \times 2 \times 1) (3 \times 2 \times 1)}$

$=56$

"he asked his mum to choose another 3 pairs out of the 5 chosen pairs at random. " So, we get

$=5C3$

$= \frac {5!}{(3!) (5-3)!}$

$= \frac {5 \times 4 \times 3 \times 2 \times 1 }{(3 \times 2 \times 1) (2 \times 1)!}$

$=10$

Finally, we get

$=56 \times 10$

$=560$

We'll take $5$ pairs from the $8$ pairs of shoes. This is $\frac {8!}{5!\cdot3!} = 56$ . For each of these $56$ combinations, we can take $3$ pairs, and this is $\frac {5!}{3!\cdot2!} = 10$ . Finally, there are $\boxed {560}$ possible ways for the $3$ pairs of shoes to be chosen at random in the end.

No. of possible ways = ( 8C3 ) * ( 5C3 ) = 560 ways

Out of 8 pair he asked his father to choose 5 pairs, therefore, no.of ways of selecting 5 pairs = 8C5 = 56

Now, he asked his mom to select another 3 pair, therefore, no. of ways = 5C3 = 10

Total no. of ways = 56 * 10 = 560

Selection of 5 objects from 8 dinstinct objects is possible in 56 ways while selection of 3 from those 5 is only possible in 10 ways. Thus giving us 56x10 possible ways for the two events happening together.

answer is 8C5 * 5C3

cause, dad can choose 5 pair of shoes in 8C5 ways

and for each combinat of ion of 5 pairs , his mother can choose 3pairs of shoes in 5C3 ways

so, ans= 8C5*5C3=560