2005 AMC 10B Problem 21

Forty slips are placed into a hat, each bearing a number

1,

2,

3,

4,

5,

6,

7,

8,

9,

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}?

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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:

\begin{aligned} \binom{10}{2}\binom{4}{2}\binom{4}{2} &= 45 \cdot 6 \cdot 6 \\ &= 1620. \end{aligned}

Therefore

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

Thus, A is the correct answer.