2005 AMC 10B Problem 21

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

Both probabilities share the denominator

\binom{40}{4},

so only the counts matter

Count four-of-a-kind draws, then draws of the form two

a

’s and two

b

’s

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.

Problem 21 in other years