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Shubhradeep
Option C should be correct

\(10^{7}\)­ = 10000000 , so, we need 8 digit numbers which has 6 as the product of their digits.

6 = 2x3
6 = 1x6


we can have S1 = {1,1,1,1,1,1,2,3} and S2 = {1,1,1,1,1,1,1,6}

We can arrange the dogits of S1 in \(\frac{8!}{6!}\) = 7x8 = 56
We can arrange the dogits of S2 in \(\frac{8!}{7!}\) = 8x1 = 8

So, total number of natural numbers = (56+8) = 64

Hence, option C­
­Hey!
Very helpful your answer but I need your help to clarify something because I'm confused. I also think it's C, but regarding the fractions, in the first one the denominator is 6! because in the first set only 2 out of the 8 numbers (2 and 3) have product equal to 6?

If it is so, then why isn't it also in the second denominator?­
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Gmatguy007

Shubhradeep
Option C should be correct

\(10^{7}\)­ = 10000000 , so, we need 8 digit numbers which has 6 as the product of their digits.

6 = 2x3
6 = 1x6


we can have S1 = {1,1,1,1,1,1,2,3} and S2 = {1,1,1,1,1,1,1,6}

We can arrange the dogits of S1 in \(\frac{8!}{6!}\) = 7x8 = 56
We can arrange the dogits of S2 in \(\frac{8!}{7!}\) = 8x1 = 8

So, total number of natural numbers = (56+8) = 64

Hence, option C­
­Hey!
Very helpful your answer but I need your help to clarify something because I'm confused. I also think it's C, but regarding the fractions, in the first one the denominator is 6! because in the first set only 2 out of the 8 numbers (2 and 3) have product equal to 6?

If it is so, then why isn't it also in the second denominator?­
­
It's not actually that way which you are assuming. It seems You are assuming that 2x3 makes 6, so 2 digits are gone that's why we make 6! in the denominator and same should be the case for the second one.

We have to find number of natural numbers which have products of their digits as 6.
We can make it this way - number of arrangements of digits possible with the factors of 6 - which is eventually 2, 3 in first case and 1,6 in second case.

Now it boils down to this problem-
Case 1: Number of possible arrangements with digits 1,1,1,1,1,1,2,3
Case 2: Number of possible arrangements with digits 1,1,1,1,1,1,1,6

I shall give an example so that you understand
Suppose, we have few letters L,E,T,T,E,R,S
How many ways you can arrange these letters( or how many words can you make with these leters)

Frequency of each letter:
L-> 1
E-> 2
T-> 2
R-> 1
S-> 1
Total 7 letters

So we can make \(\frac{7!}{(1!*2!*2!*1!*1!)}\) = \(\frac{7!}{(2!*2!)}\) words

This 2!x2! in the denominator is for the repeating letters E and T. So If we would have written 7! possible arrangements possible that would have been wrong because  'EE' or 'TT' is 1 combination, so in each case we have to divide by 2!

Same is with the cases in the given question.
There are 6 repeating 1's in the first case so divide by 6! and in the second case 7 repeating 1's in the second case so divide by 7!
 ­­­­­
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Shubhradeep

Gmatguy007

Shubhradeep
Option C should be correct

\(10^{7}\)­ = 10000000 , so, we need 8 digit numbers which has 6 as the product of their digits.

6 = 2x3
6 = 1x6


we can have S1 = {1,1,1,1,1,1,2,3} and S2 = {1,1,1,1,1,1,1,6}

We can arrange the dogits of S1 in \(\frac{8!}{6!}\) = 7x8 = 56
We can arrange the dogits of S2 in \(\frac{8!}{7!}\) = 8x1 = 8

So, total number of natural numbers = (56+8) = 64

Hence, option C­
­Hey!
Very helpful your answer but I need your help to clarify something because I'm confused. I also think it's C, but regarding the fractions, in the first one the denominator is 6! because in the first set only 2 out of the 8 numbers (2 and 3) have product equal to 6?

If it is so, then why isn't it also in the second denominator?­
­
It's not actually that way which you are assuming. It seems You are assuming that 2x3 makes 6, so 2 digits are gone that's why we make 2! in the denominator and same should be the case for the second one.

We have to find number of natural numbers which have products of their digits as 6.
We can make it this way - number of arrangements of digits possible with the factors of 6 - which is eventually 2, 3 in first case and 1,6 in second case.

Now it boils down to this problem-
Case 1: Number of possible arrangements with digits 1,1,1,1,1,1,2,3
Case 2: Number of possible arrangements with digits 1,1,1,1,1,1,1,6

I shall give an example so that you understand
Suppose, we have few letters L,E,T,T,E,R,S
How many ways you can arrange these letters( or how many words can you make with these leters)

Frequency of each letter:
L-> 1
E-> 2
T-> 2
R-> 1
S-> 1
Total 7 letters

So we can make \(\frac{7!}{(1!*2!*2!*1!*1!)}\) = \(\frac{7!}{(2!*2!)}\) words

This 2!x2! in the denominator is for the repeating letters E and T. So If we would have written 7! possible arrangements possible that would have been wrong because  'EE' or 'TT' is 1 combination, so in each case we have to divide by 2!

Same is with the cases in the given question.
There are 6 repeating 1's in the first case so divide by 6! and in the second case 7 repeating 1's in the first case so divide by 7!
 ­
­
Yeah, I was assuming that since 2*3 = 6, so 2 digits are gone that's why we make 6! in the denominator and same should be the case for the second one. But through the example it's now crystal clear, thanks Shubhradeep
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