First slide

Permutations

Question

Let m and n be two digit natural numbers. The number of pairs $(m, n)$ such that $n$ can be subtracted from m without borrowing is:

Moderate

A

2475
B

2550
C

2675
D

2875

Solution

Suppose $m = 10 a + b$ and $n = 10 c + d$ where $1 \leq a, c$ $\leq 9$ and $0 \leq b, d \leq 9 .$
Note that $0 \leq d \leq b$ Thus $d$ can take $(b + 1)$ values.
Similarly, $1 \leq c \leq a,$ therefor, $c$ can take a values.
Hence, the required number of pairs $(m, n)$ is $(1 + 2 + \dots + 10) (1 + 2 + \dots + 9) = 2475$

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