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A number is chosen at random from the set of consecutive natural numbers $\{1, 2, 3, \ldots, 24\}$. What is the probability that the number chosen is a factor of $4!$? Express your answer as a common fraction.

 Apr 6, 2020
 #1
avatar+34 
+2

\(\)4!=24

 

The factors of 24 are 1,2,3,4,6,8,12,24 and there are 8 of them

 

Therefore there is an \(\frac{8}{24}\) or \(\frac{1}{3}\) chance that the number chosen is a factor of 4!

 

Hope this helps!

-Ako

 Apr 6, 2020
 #2
avatar+930 
+1

In 4 factorial , it is 1*2*3*4 which factorizes to 2^3*3. This means is has 8 factors: 1,2,3,4,6,8,12, and 24. If a number is randomly chosen from 1-24, 8 out of the 24 would be factors of 4!. That means your answer will be 8/24=1/3.

 

 

Hope it helps!

 Apr 6, 2020
 #3
avatar+392 
+1

thx both of u

 Apr 6, 2020

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