in how many possible ways you can write 3240 as product of 3 positive integers a b c a 450

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??
10 years agoHelpfull: Yes(21) No(4)
how u got 5c2,6c2,3c2 @SARASWATHI
10 years agoHelpfull: Yes(6) No(1)
hw can u write 5c2, x1+x2+x3=3 k explain plz
10 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
10 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
10 years agoHelpfull: Yes(2) No(0)
3240=2*2*2*5*3*3*3*3

so, the answer is 8!/(3!*4!)
10 years agoHelpfull: Yes(1) No(9)
sorry the above one is the wrong solution.
10 years agoHelpfull: Yes(1) No(2)
how did u get 6c2 and 3c2
make it clear @saraswathy
10 years agoHelpfull: Yes(1) No(1)
2^3*3^4*5^1=3240.....1/2(3+1)(1+1)(4+1)=20
10 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)

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