Algebra - find the number of all postive integral solutions

49

what we have to do is find out ways of choosing 3 distinct factors of the given thing

62

yes that is not the solution.. because this is diferent from partitioning

here the powers of 2 have to partitioned seperately, then power of 3 and then the power of 5..

so your answer should be 4C2 x 4C2 x 5C2

49

nishant bhaiya bt i wanted to post the answer

62

see for each power of 2 there are options that the powers can go to either x1 or x2 or x3

so we are partitioning each power seperately..

@shubhomoy: sorry i replied before

1

The first way i'm showing...the rest you can make out.

Divide 2² into the factors of x1, x2, x3 (partitionning).

n = 2 (as no. of factors of 2² = 2 here )

r = 3( x1, x2, x3)

Hence no. of ways = n+r-1 C r-1 = 4C2

for 3² : 4C2

For 5³ : n = 3 (5,5,5) ; r = 3 (x1, x2, x3) So 5C2

Hence no. of solns. = the total no . of ways of all these events occurring simultaneously : 4C2 x 4C2 x 5C2

If you've understood...plzzzz reply

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find the number of all postive integral solutions