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2005 AMC 10B Problem 21

2005 AMC 10B Problem 21AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Probability · difficulty 4 of 5 · about 4 min · position P21-25

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 $q/p$ ?

A)
$162$
B)
$180$
C)
$324$
D)
$360$
E)
$720$

Answer

Key insight

Both probabilities share the denominator C(40,4), so q/p is just a ratio of favorable counts: 45 pairs of numbers times 6 times 6, over 10.

Solution

Treat a draw as an unordered set of $4$ slips out of $40$ ; every such set is equally likely, and there are $\binom{40}{4}$ of them. Since $p$ and $q$ share this denominator, $q/p$ equals the ratio of the numbers of favorable sets.

Four of a kind. Pick the number ( $10$ ways); all four of its slips are then taken. Favorable sets: $10$ .

Two pairs. Pick the two distinct numbers as an unordered pair: $\binom{10}{2} = 45$ ways. For each chosen number, pick which $2$ of its $4$ slips are drawn: $\binom{4}{2} = 6$ ways each. Favorable sets: $45 \cdot 6 \cdot 6 = 1620$ .

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

The answer is $\boxed{\textbf{(A)}\ 162}$ .

Why this works

When two probabilities are defined on the same experiment, their ratio never needs the total; only the favorable counts matter. Counting unordered hands keeps the bookkeeping minimal, and the structure "choose the values, then choose the slips for each value" is the standard template for poker-style counts.

Alternative approach

Ordered draws work too: $p$ has $10 \cdot 4!$ favorable sequences and $q$ has $45 \cdot \binom{4}{2} \cdot (4 \cdot 3)(4 \cdot 3)$ , and the ratio is again $\frac{45 \cdot 6 \cdot 144}{240} = 162$ . Mixing ordered and unordered counts between $p$ and $q$ is the main hazard.

The trap

Counting the two numbers a and b as an ordered pair (10 times 9 = 90 instead of 45), which doubles the answer to 324.

Common mistakes

Counting the two numbers a and b as an ordered pair (10 times 9 = 90 instead of 45), which doubles the answer to 324.
Forgetting the $\binom{4}{2}$ choices of slips for one of the two numbers, giving $45 \cdot 6 / 10 = 27$ or similar.

Techniques

Set up the equation/formula and compute; no special trick needed