Question

Total number of integral solution of (a, b, c) such that abc = 105 is

• A. 6
• B. $3P_{3}$
• C. Equivalent to total number of integral solution of abc = 210
• D. Equivalent to total number of integral solution of abc = 420

Answer 1

Answer: (C)

Equivalent to total number of integral solution of abc

= 210

Answer 2

abc = 105 = 3 x 5 x 7

∴ Required ways = 3 ! = 6 = $3P_{3}$ if abc = 210 = 2 x 3 x 7 x 5

We can choose different triplets in $4C_{2}$ ways each triplet can be permutated in 3! Ways.

Thus number of integral solution of abc = 210 is $4C_{2}\mspace{6mu} x\mspace{6mu} 3!$.