We need to find the probability of getting 3 queens and 2 kings
In 52 cards, we have 4 queens and 4 kings
We need to select 3 queens from 4 queens
\[\frac{n!}{r!(n-r)!}\]
So different combinations of selecting 3 queens from 4 queens is 4C3
\[\frac{4!}{3!(4-3)!}\] = 4
We need to select 2 kings from 4 kings
\[\frac{n!}{r!(n-r)!}\]
So different combinations of selecting 2 kings from 4 kings is 4C2
\[\frac{4!}{2!(4-2)!}\] = 6
Total combination of selecting 5 cards from 52 cards = 52C5
\[\frac{52!}{5!(52-5)!}\] = 2598960
Probability of getting 3 queens and 2 kings = 4C3 * 4C2 / 5C52
\[\frac{4*6}{2598960}\]
1/ 108290