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

• 3240=2*2*2*3*3*3*3*5;
  =8!/3!*4!;
  280
• 9 years agoHelpfull: Yes(3) No(3)
• 3240 = 2*2*2*3*3*3*3*5
  
  we have the formula to arrange 3+ve integers as (n+r-1) C (r - 1)
  
  so arranging 2 as ------> (3+3-1) C (3-1) = 10 ways
  
  3 as ------> (3+4-1) C (3-1) = 15 ways
  
  5 as ------> 3 ways
  
  therefore total no of possible ways of writing 3240 as product of 3+ve integers = 10 * 15 * 3
  = 450 ways
• 9 years agoHelpfull: Yes(3) No(0)
• the factors of 3240 are 5,2,3,3,3,2
  so the no of ways are (6*5*4)/3!2!1!=10
  Becoz we are arranging the values in 6*5*4 but some value are repeated so we have to divide these values so the answer will be 10
• 10 years agoHelpfull: Yes(2) No(24)
• there can also be combined no. at each position....
  so true answer is still ambiguous
• 10 years agoHelpfull: Yes(0) No(5)
• 60 ways! 3240 has factors- 2,2,2,9,9,5
  therefore soln is :- 6!/2!*3!=60 ways
• 10 years agoHelpfull: Yes(0) No(4)