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Q.
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:
Permutations and Combinations
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, $1 \leq c \leq a$, therefore, $c$ can take $a$ values. Hence, the required number of pairs $(m, n)$ is $(1+2+\ldots+10)(1+2+\ldots+9)=2475$