# 1995 AHSME Problems/Problem 11

## Problem

How many base 10 four-digit numbers, $N = \underline{a} \underline{b} \underline{c} \underline{d}$, satisfy all three of the following conditions?

(i) $4,000 \leq N < 6,000;$ (ii) $N$ is a multiple of 5; (iii) $3 \leq b < c \leq 6$. $\mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 18 } \qquad \mathrm{(C) \ 24 } \qquad \mathrm{(D) \ 36 } \qquad \mathrm{(E) \ 48 }$

## Solution

• For condition (i), the restriction is put on $a$; $N<4000$ if $a<4$, and $N \ge 6$ if $a \ge 6$. Therefore, $a=4,5$.
• For condition (ii), the restriction is put on $d$; it must be a multiple of $5$. Therefore, $d=0,5$.
• For condition (iii), the restriction is put on $b$ and $c$. The possible ordered pairs of $b$ and $c$ are $(3,4)$, $(3,5)$, $(3,6)$, $(4,5), (4,6),$ and $(5,6),$ and there are $6$ of them. Alternatively, we are picking from the four digits 3, 4, 5, 6, and for every combination of two, there is exactly one way to arrange them in increasing order, so we have $\binom{4}{2} = 6$ choices for $b$ and $c$ when we consider them together.

Multiplying the possibilities for each restriction, $2 \cdot 2 \cdot 6=24\Rightarrow \mathrm{(C)}$.

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