in how many ways can 2310 be expressed as a product of three factors

• 2130=2*5*7*11*3*1
  6 prime factors...to express this as a product of 3 we take, (n-1)c(r-1)
  5c2=10
• 10 years agoHelpfull: Yes(16) No(15)
• ans is (41)
• 10 years agoHelpfull: Yes(6) No(1)
• 2310 = 2*3*5*7*11
  We have to form a group of 3 out of these 5, it can be done in
  5C2*3C2+5C3 ways i.e. 40 ways
• 10 years agoHelpfull: Yes(4) No(9)
• three is right answer??
  
• 10 years agoHelpfull: Yes(3) No(4)
• Is The ans 60
• 10 years agoHelpfull: Yes(1) No(4)
• 2130 has 5 factor that are = 2,5,3,711;
  so ways of making product of three factors= 5c3=5c2=10
• 8 years agoHelpfull: Yes(1) No(0)
• in 375 ways !!!
• 10 years agoHelpfull: Yes(0) No(0)
• 2310 = 2*3*11*5*7
  therefore, x1+x2+x3+x4+x5 = 5
  (5+3-1)C(3-1) = 21
• 8 years agoHelpfull: Yes(0) No(0)
• answer : 81.
  2130 = 2*5*7*11*3
  to experess this as a product of 3 factors let it be abc
  2(only 1 digit & it it is to be counted 2 didits if it is 2^2) distributed to these 3 digits in [(1+3)C(3-1)]=3C2 ways = 3 ways
  similary for 5,7,11,3 for each factor there will be 3 ways
  hence total number of ways = 3*3*3*3*3= 243 ways.
  Ans : 243 ways
• 8 years agoHelpfull: Yes(0) No(0)