Question

Forty slips are placed into a hat, each bearing a number $1, 2, 3, 4, 5, 6, 7, 8, 9,$ or $10,$ with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b \ne a.$ What is the value of $\dfrac{q}{p}?$

$162$

$180$

$324$

$360$

$720$

Solution:

Both events draw from $\binom{40}{4}$ equally likely selections, so $\dfrac{q}{p}$ is the ratio of their favorable counts.

Exactly $10$ draws give four slips of the same number, one for each value.

For two $a$ 's and two $b$ 's, choose the two values in $\binom{10}{2}$ ways, then two of the four $a$ -slips and two of the four $b$ -slips: $\binom{10}{2}\binom{4}{2}\binom{4}{2} = 45 \cdot 6 \cdot 6 = 1620.$

Therefore $\dfrac{q}{p} = \dfrac{1620}{10} = 162.$

Thus, A is the correct answer.

2005 AMC 10B Problem 21

Solution:

Problem 21 in Other Years