What is the number of ways of expressing 3600 as a product of three ordered positive integers (abc, bca etc. are counted as distinct). For example, the number 12 can be expressed as a product of three ordered positive integers in 18 different ways.
a. 441
b. 540
c. 84
d. 2100

• 3600 = 2^4 × 3^2 × 5^2
  Let abc = 2^4 × 3^2 × 5^2
  We have to distribute four 2's to three numbers a, b, c in (4+3−1)C(3−1)=6C2 = 15 ways.
  Now two 3's has to be distributed to three numbers in (2+3−1)C(3−1)=4C2 = 6 ways
  Now two 5's has to be distributed to three numbers in (2+3−1)C(3−1)=4C2 = 6 ways
  Total ways = 15 × 6 × 6 = 540
• 8 years agoHelpfull: Yes(23) No(1)
• 3600=2^4*3^2*5^2... We have to form abc.. So it can be formed by the formula n+r-1c r-1. So total ways 4+3-1c3-1 * 2+3-1c3-1 * 2+3-1c3-1
  As we have to make a form of abc , so r=3.
  Hence 6c2 * 4c2 * 4c2 = 540.
• 8 years agoHelpfull: Yes(5) No(3)
• answer is 540
• 8 years agoHelpfull: Yes(1) No(2)