For any integer P greater than 1, P! denotes the product of all intege

Step 1: Found the Total No. of Factors in 8!

the Prime Factorization of 8! = 2^7 * 3^2 * 5^1 * 7^1

Total No. of +Positive Factors = 8 * 3 * 2 * 2 = 96 Total + Pos. Factors


Step 2: Probability = No. of Fav. Outcomes in which the Factor includes 2^2 * 7^1 / 96 Total Factors


In order for 1 of the Factors to be a Multiple of 28, it must include: 2^2 and 7^1

this leaves the Prime Factors of: 2^5 * 3^2 * 5^1


How many different Combinations can we have for each Prime Factor Above given that the 0th Power can be included for each (if they are all to the 0th Power, then the Factor just = 2^2 * 7^1 = 28). All these different Combinations can be combined with 2^2 and 7^1 to create Multiples of 28

How many Options for the Prime Factor 2?

6 options

How many Options for the Prime Factor 3?

3 options

How many Options for the Prime Factor 5?

2 options

6 * 3 * 2 = 36 Total Multiples


Answer: Probability = 36 / 96 = 3/8

Answer C


EDIT: just saw the post above mine. Same exact concept. Sorry lol

We have to find the factors of 8! and \(\frac{8!}{28}\), and to find the factors we have to get the numbers in their prime factorization.

8! would consist of all the prime factors till 8, so would contain 2, 3, 5, 7 and their powers.
\(8!=2^a*3^b*5^c*7^d\)
Factors of 2 in 8! =[\(\frac{8}{2}\)]+[\(\frac{8}{2^2}\)]+[\(\frac{8}{8}\)], where [a] gives the integer value less than or equal to a.
Factors of 2 = 4+2+1=7
Factors of 3 in 8! =[\(\frac{8}{3}\)]+[\(\frac{8}{3^2}\)]=2+0=2
Factors of 5 in 8! =[\(\frac{8}{5}\)]+[\(\frac{8}{5^2}\)]=1+0=1
Factors of 7 in 8! =[\(\frac{8}{7}\)]+[\(\frac{8}{7^2}\)]=1+0=1
So \(8!=2^7*3^2*5*7\), where number of factors =\((7+1)*(2+1)*(1+1)*(1+1)=8*3*2*2\)

Now \(\frac{8!}{28}\) would be \(2^5*3^2*5^1\), where number of factors =\((5+1)*(2+1)*(1+1)=6*3*2\)

Probability = \(\frac{6*3*2}{8*3*2*2}=\frac{3}{8}\)


C

8! = (2^7)(3^2)(5)(7)

To find a random factor of 8!, we can generate a number that looks like this: 2^a * 3^b * 5^c * 7^d, where a can take any value from 0 to 7 inclusive, b any value from 0 to 2 inclusive, and c and d can be 0 or 1. If we want our random factor to be a multiple of 2^2*7, we need d to equal 1, which will happen 1/2 the time. We also need a to be at least 2, which will happen 6/8 of the time (six of the eight possible values of a are 2 or greater). We don't care about b or c. So the answer is (1/2)(6/8) = 3/8.

edited - i originally had a typo at the very end

Why do you use the remaining factors instead of its factors itself????
Thank you in advance