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:

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: (A) 2475 (B) 2550 (C) 2675 (D) 2875

Tardigrade Admin · Accepted Answer

Suppose m=10 a+b and n=10 c+d where 1 ≤ a, c ≤ 9 and 0 ≤ b, d ≤ 9. Note that 0 ≤ d ≤ b. Thus d can take (b+1) values. Similarly, 1 ≤ c ≤ 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

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:

2475

2550

2675

2875

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

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: