in how many possible ways you can write 3240 as product of 3 positive integers a, b, c .
a 450
b 420
c 350
d 320

• 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
  
  Ans : (A) 450
• 9 years agoHelpfull: Yes(27) No(9)
• X1+X2+X3=3=>5C2
  because x1+x2+x3----xr=n =>(n+r-1)C(r-1),
  Here n=3 and r=3
  so, put n and r in formula,we get 5C2.
  Is it clear now??
• 9 years agoHelpfull: Yes(21) No(4)
• how u got 5c2,6c2,3c2 @SARASWATHI
• 9 years agoHelpfull: Yes(6) No(1)
• hw can u write 5c2, x1+x2+x3=3 k explain plz
• 9 years agoHelpfull: Yes(2) No(4)
• Continuation of saraswathi
  we get x1+x2+x3=3=>this can be satisfied in 10 ways,similarly the other two can be satisfied in 12 and 3 ways.so intotal the answer is 10*15*3=450
• 9 years agoHelpfull: Yes(2) No(0)
• no of positive solns for a some nos ie x1+x2+x3=3 is given by a formula.this formula gives the non neg sols for such kind of eqns....
  formula (3+3-1)c(3-1)this gives 5c2...this is actually done by this saraswati...
  this is the key for those who dont know the formula
• 9 years agoHelpfull: Yes(2) No(0)
• 3240=2*2*2*5*3*3*3*3
  
  so, the answer is 8!/(3!*4!)
• 9 years agoHelpfull: Yes(1) No(9)
• sorry the above one is the wrong solution.
• 9 years agoHelpfull: Yes(1) No(2)
• how did u get 6c2 and 3c2
  make it clear @saraswathy
• 9 years agoHelpfull: Yes(1) No(1)
• 2^3*3^4*5^1=3240.....1/2(3+1)(1+1)(4+1)=20
  
• 9 years agoHelpfull: Yes(0) No(3)
• By lcm, we calculate 3240=2^3 * 3^4 * 5
  as a product of 3 integers a*b*c
  we arrange 2^3 in 10 ways
  and 3^4 in 15 ways
  nd 5 in 3 ways
  totally 10*15*3 = 450 ways
• 8 years agoHelpfull: Yes(0) No(1)