In how many ways in which [m]5^7[/m] can be expressed as a product of

Question

In how many ways in which  5^7  can be expressed as a product of 3 positive factors?

A. 6
B. 7
C. 8
D. 9
E. 10

chetan2u · Answer

So  5^7=5^{a+b+c} ..
So we look for values of a,b,c when a+b+c=7, and a can be 0 too as 5^0=1, a positive integer.
There are straight formulas, but GMAT does not test that, so we should be able to calculate all ways as below.
Let us work for different values of a..
1) a=0 and b+c=7
(b,c)=(0,7), (1,6),(2,5),(3,4)-------- 4 ways
2) a=1 and b+c=6, where none of B and c are 0 as they have been calculated in case 1.
(b,c)= (1,5),(2,4),(3,3)-------- 3 ways
3) a=2 and b+c=5
(b,c)=(2,3)-------- 1 way
No other possibilities

Total 4-3+1=8

C

masoodtaxed · Answer

What if a =3 and b=2 c=2 that will also be 7 right?

nick1816 · Answer

Because we looking for Unordered solutions of (a,b,c), order of a,b and c doesn't matter.

Neha2050 · Answer

for third case there is one more possibility
3) a=2 and b+c=5
(b,c)=(2,3)-------- 1 way
b,c - 4,1

resulting in total 9 ways
kindly clarify if wrong

nick1816 · Answer

He included that in the 2nd case. (1,2,4) is similar to (2,4,1), as we are finding the un-ordered solution.

ShaikhMoice · Answer

Thank you chetan2u for the explanation.
I was trying to solve the question using the formula
a+b+c=7
No of ways to distribute =  (7+3-1) C (3-1) = 36 ways
To remove arrangement divided the answer by 3!
Final answer is coming to be 6.
Could you please explain where i am commiting the mistake.
Thanks in advance

Fdambro294 · Answer

(5)^7 can only have factors of 5 that have different exponent values that, in the end, sum to 7

For example, to have 3 positive factors multiply to (5)^7 we could have:

(5)^1 * (5)^4 * (5)^2 = (5)^7

Also, (5)^0 = 1 would be a factor so we could have something like:

(5)^0 * (5)^3 * (5)^4 = (5)^7

Since the rules for exponents say we add the exponents when multiplying terms with the same base and the ordering of the factors does not matter when we multiply —-

Essentially, the problem boils down to Distributing 7 Identical items (the 7 “1’s” that make up the sum of 7) ——-> into 7 IDENTICAL “bins”

In this type of problem it does not matter which object goes into a “bin”. All the matters is how many objects goes into each “bin”. In addition, the ordering of these “bins” does not matter.

Any bin can have 0 and any bin can have all 7, so long as the amounts in three identical bins sum to 7.

(Formally, this problem is akin to partitioning an integer)

0 + 0 + 7 
0 + 1 + 6
0 + 2 + 5
1 + 1 + 5
0 + 3 + 4
1 + 2 + 4
0 + 3 + 3
1 + 2 + 3

8 possibilities

Answer

8

Posted from my mobile device

mpobisetty · Answer

Hi  chetan2u , can you let me know what formula can be used here?

As per my understanding, number of ways of distributing n identical objects in r distinct boxes such that each box is filled with 0 or more objects is C(n+r-1,r-1).

Here, n = 7, r=3. But this doesn't give the right answer.

IanStewart · Answer

That formula is for ordered partitions: it would give the right answer if you wanted to think of 1, 2, 4 and 2, 1, 4, say, as two different answers. From the wording of this question, it's not clear whether we want to count ordered partitions or unordered ones, but from the answer choices, the question intends for order not to matter (so we're meant to think of (5^1)(5^2)(5^4) and (5^2)(5^1)(5^4) as if they were the same). So the formula you've tried to use will give an answer that is much too large. There is actually no convenient formula for unordered partitions (there is a formula of course, but it's something called a 'recurrence relation', where you need to sum the answers you'd get using small numbers to find the answer with bigger numbers), and regardless, if you're studying for the GMAT, you'd never need such a formula anyway. I've never even seen a standard partition problem on the test, let alone an unordered partition problem.

chetan2u · Answer

Hi,

This would also be a response to post of  ShaikhMoice

The formula for a+b+c=7 is (7+3-1)C(3-1), which is 36.

Now you cannot divide the entire ways by 3!, as not each case has 3! of arrangements. For example : (0,0,7) or (1,1,5) will have only 3 ways of arrangements.

So, how do we move ahead?
Subtract the ways in which two or all three are same. 
All 3 same: no way as 7 is not divisible by 3.
Only two same: Each of these same numbers would be 0 to 3 as 4 will make total 4*2 or 8 >7.

When two are same, the ways 3 can be arranged are 3!/2! or 3 ways. 
4 such numbers (0 to 3) will give 4*3 12 ways.

Finally ways with all different numbers = 36-12 = 24.
Dividing this by 3!, we get 24/6 or 4.

Also we had got 4 ways where two are same.

Thus 4+4 or 8 ways.

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In how many ways in which [m]5^7[/m] can be expressed as a product of

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