find no. of solutions

Not at all my doubt

Find total no. of positive integral solutions of xyz = 23.34.51

16 Answers

106
Asish Mahapatra ·

??

1
newmember ·

is it (3C1 + 3C2.2! + 3C3).(3C1 + 3C2.2! + 3C2+3C3.3!/2!).3C1 ??

1
Philip Calvert ·

38

edit : This is definitely wrong.See eureka and Nishant bhaiyyas posts below

1
fibonacci ·

i think philip sir is correct

24
eureka123 ·

no. of ways=(^5C_2)(^6C_2)(3)

24
eureka123 ·

Explaianation:
No of ways to distribute 3 identical 2's =^5C_2
no. .of ways to distribute 4 identical 3's =^6C_2
no .of ways to diribute 1 5 =3

1
fibonacci ·

i can explain philip sir's answer
as each of the prime factors have a choice of going to 3 places ie to x, to y, to z . and there are 8 prime factors(need not be distinct)
so ans comes out to be 3*3*3*3*3*3*3*3=38
am i right philip sir

24
eureka123 ·

I dont think philip's ans is right

1
Philip Calvert ·

@fibonacci : first of all delete that "sir" :( :(

@eureka : maybe.

I just wrote what came to my mind just then.

24
eureka123 ·

[3]

1
Arshad ~Died~ ·

phillip sir [3]
can u tell me how u got 38

62
Lokesh Verma ·

38 is not the right answer.. there is a bit of repetition in that

I think the right answer is number of ways to divide 3 in 3 parts times 4 in 3 parts times 1 in 3 parts

so the answer is

5C2x6C2x3C2

Which is the answer given by eureka :)

24
eureka123 ·

tahts what i wrote in post #7......but hid it so that everyone could try..[1]

its ball and stick method...

106
Asish Mahapatra ·

@newmember: your answer is perfectly fine.... but Eureka's method is much more easier to understand...

Thanks Eureka and Nishant sir..

11
virang1 Jhaveri ·

eureka can u explain a bit more pls

24
eureka123 ·

@virang
see teh hidden part in #7

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