In how many possible ways can you write 3240 as a product of 3 positive integers a, b and c.

• COULD U EXPLAIN KAUSHIK HOW YOU SOLVE THAT
• 6 years agoHelpfull: Yes(2) No(0)
• Make the given number as prime factors.,(Ex:3240 as
  324*10
  2*162 5*2(for 10)
  2*81
  3*3*3*3
  so total 2^3 3^4 5^1
  we have formula. That is (n+r-1)C(r-1)
  where n is power and r is no.of positive integers
  by that, for 2^3 we get 10 and for 3^4 we get 15 and finally for 5^1 we get 3
  by multiplying all 10*15*3=450
  ans is 450. :)
• 6 years agoHelpfull: Yes(2) No(0)
• First do prime factorization for 3240.
  2^3*5*3^4
  multiply 2*5+5*3+3=10*15*3=450
• 6 years agoHelpfull: Yes(1) No(2)
• 450=5c2* 6c2*3C2
• 6 years agoHelpfull: Yes(0) No(9)
• please, Anybody tell guys how it comes.....
• 6 years agoHelpfull: Yes(0) No(1)
• 5c2*6c2*3c2=450
• 6 years agoHelpfull: Yes(0) No(1)
• there are 3 no(condition not given for repetation)
  _ _ _
  inthe last no maybe 2,3,4 so 3 posible
  inthe mid no maybe 2 3 4 so 3 posible
  the first no may be 02 03 04,20,30,40,22,23,24,33,32,34,42,43,44 there 15 posible
  so answer is 3*3*15=135
• 6 years agoHelpfull: Yes(0) No(1)