In how many possible wasy can write 3240 as a product of 3 positive integers a,b and c.
Option
a) 450
b)420
c)350
d)320

• hey help me......say the ans how 8!/3!4!1!=420 but it comes 280 only ...........
  
• 9 years agoHelpfull: Yes(6) No(2)
• 3240 => 2^3 * 3^4 * 5^1
  a => 2^x1 * 3^y1 * 5^z1
  b => 2^x2 * 3^y2 * 5^z2
  c => 2^x3 * 3^y3 * 5^z3
  a * b * c = 3240
  x1 + x2 + x3 = 3 => 5C2
  y1 + y2 + y3 = 4 => 6C2
  z1 + z2 + z3 = 1 => 3C2
  5C2 * 6C2 * 3C2 = 10 * 15 * 3 = 450
  anwer=450
• 9 years agoHelpfull: Yes(6) No(1)
• Answer is 450
• 9 years agoHelpfull: Yes(3) No(0)
• 3240= 2*2*2*5*3*3*3*3
  8!/3!4!1!=420
  so ans=420
• 9 years agoHelpfull: Yes(2) No(7)
• can anyone explain what method we will use
• 10 years agoHelpfull: Yes(1) No(0)
• can you explain your solution @GOPINATH. how did you write 5c2 , 6c2,...
• 9 years agoHelpfull: Yes(1) No(0)
• http://www.m4maths.com/33219-In-how-many-possible-ways-can-write-3240-as-a-product-of-3-positive-integers-a-b-and-c.html
• 10 years agoHelpfull: Yes(0) No(20)