2005 AMC 10B Problems/Problem 21

Problem

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 $b \neq a$ . What is the value of $q/p$ ?

$\mathrm{(A)} 162 \qquad \mathrm{(B)} 180 \qquad \mathrm{(C)} 324 \qquad \mathrm{(D)} 360 \qquad \mathrm{(E)} 720$

Solution (where the order of drawing slips matters)

There are $10$ ways to determine which number to pick. There are $4!$ way to then draw those four slips with that number, and $40 \cdot 39 \cdot 38 \cdot 37$ total ways to draw four slips. Thus $p = \frac{10\cdot 4!}{40 \cdot 39 \cdot 38 \cdot 37}$ .

There are ${10 \choose 2} = 45$ ways to determine which two numbers to pick for the second probability. There are ${4 \choose 2} = 6$ ways to arrange the order which we draw the non-equal slips, and in each order there are $4 \times 3 \times 4 \times 3$ ways to pick the slips, so $q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \times 39 \times 38 \times 37}$ .

Hence, the answer is $\frac{q}{p} = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{10\cdot 4!} = \boxed{\ \mathbf{(A)}162}$ .

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2005 AMC 10B Problems/Problem 21

Problem

Solution (where the order of drawing slips matters)

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