IIT JEE , AIEEE Forum - Mathematics , Algebra - applications of permutations and combinations

aakash_128

Find the number of factors(excluding 1 & the expression itself)of the product of

a^7 b^4 c^3 d e f where a,b,c,d,e,f are all prime numbers....

^ stands for to the power....

SIR,

Please help me sort out this challenging question....

Thank you.......

magiclko

Hiiiiiiiiii

Well its quite simple aakash,

u mst hav read that no of ways of selecting none or more objects from n identical objects = n+1

so suppose m= a^p b^q c^r, whr all a,b and c are prime nos,

we are having total no of factors = (p+1) (q+1) (r+1)

hopefully this wuld be clear to u, if not reply back!!!

magiclko

i think it will be a li'l more clear by an e.g

if we hav to find no of factors of 72,

then we hav 72= (2^3) X (3^2)

so no of ways of selecting 2 from three 2s, such that none or more 2 is taken = 4

similarly for 3, we hav 4 ways,

so total no of ways = 4X3 = 12

krish

My dear friend aakash,in ur question
We take into consideration all prime factors a,b,c, d,e,f;so selecting them is also necessary....as far as i know , all possible ways to select factors which are repeating is n+1 where is no.of repetitions and those to select different ones is 2^r where r is the no. of different ones.....
Total ways = (p+1)(q+1)(r+1)2^(3)
On excluding 1 and the number we get
no. of factors = (p+1)(q+1)(r+1)2^(3) - 2
i think i must be correct and clear.....plz reply if i am wrong

krishna.gopal

Mansi you forgot to take 1 and the number itself out of factors of 72.

Aakash I think Krish has given you right answer but by using p,q r instead of 7,4,3

So your answer will be (7+1)*(4+1)*(3+1)*2^3-2

magiclko

yeah i forgot to exclude those two

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