Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

yrelhan4

  • 3 years ago

i have 2 questions of similar type. looking for a permutation and combination method to it Number of integral solutions of xyz=30 and xyz=24? also explain what if it was positive integral solutions? a start would be good.

  • This Question is Closed
  1. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    30 = 2 * 3 *5 24= 2 * 2 *2 * 3 If that helps ?

  2. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 is also there.

  3. yrelhan4
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thats what i did too. what next?

  4. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    factors of 30 = 1,2,3,5,6,10,15,30 Lets fix x=1 and say x<=y<=z now (y,z) can be (1,30) (2,15) (3,10) (5,6) No more. If we fix x=2, (x<=y<=z) (y,z) = (3,5) No more If we fix x=3, (x<=y<=z) (y,z) =None So we come to a halt Now note that x,y,z are all symmetrical, all x,y,z can be permuted in 3! ways except a special case when 2 are equal, then no. of permutations will be 3!/2! Ans = (4*3!) + (1* 3!/2!) =27 Unless I'd have done something wrong ?

  5. shubhamsrg
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Similar approach for 24

  6. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy