$n∗$ denotes the product of all the integers from 1 to n, inclusive.
so it is $n!$

We are asked to find the prime no between $7 ∗ +2$ and $7 ∗ +7$, inclusive.
Prime nos between $7! + 2$ & $7! + 7$

The factorial means that the no can be divisible by itself of lesser than the no. And by adding 2 to 7, they will be divisible by the same no as well.

$7! = 5040$
If we add 2, then the no will be divisible by 2
If we add 3, then the no will be divisible by 3
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.
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same goes for upto 7 is added. Hence there are no prime numbers. NONE

Answer A

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1*2*3*4*5*6*7 +2 = 2(1*3*4*5*6*7+1) --> 2 factors (Not Prime)
1*2*3*4*5*6*7 +3 = 3(1*2*4*5*6*7+1) --> 2 factors (Not Prime)
1*2*3*4*5*6*7 +4 = 4(1*2*3*5*6*7+1) --> 2 factors (Not Prime)
1*2*3*4*5*6*7 +5 = 5(1*2*3*4*6*7+1) --> 2 factors (Not Prime)
1*2*3*4*5*6*7 +6 = 6(1*2*3*4*5*7+1) --> 2 factors (Not Prime)
1*2*3*4*5*6*7 +7 = 7(1*2*3*4*5*6+1) --> 2 factors (Not Prime)

Hence no prime present.

Given that $n∗$ denotes the product of all the integers from 1 to n, inclusive and we need to find how many prime numbers are there between $7∗$ + 2 and $7∗$ + 7, inclusive

$7∗$ = Product of all the integers from 1 to 7, inclusive = 1*2*3*4*5*6*7 = 7!

$7∗$ + 2 = 7! + 2 = 1*2*3*4*5*6*7 + 2 = 2*(1*3*4*5*6*7 + 1) = a Multiple of 2 => NOT a Prime Number
Similarly, 7! is also a multiple of all numbers from 3 to 7
=> all of the numbers from 7! + 2 to 7! + 7 will be Non-Prime numbers.

So, Answer will be A.
Hope it helps!

(Working below)

$7∗$ + 3 = 7! + 3 = 1*2*3*4*5*6*7 + 3 = a Multiple of 3 => NOT a Prime Number
$7∗$ + 4 = 7! + 4 = 1*2*3*4*5*6*7 + 4 = a Multiple of 4 => NOT a Prime Number
$7∗$ + 5 = 7! + 5 = 1*2*3*4*5*6*7 + 5 = a Multiple of 5 => NOT a Prime Number
$7∗$ + 6 = 7! + 6 = 1*2*3*4*5*6*7 + 6 = a Multiple of 6 => NOT a Prime Number
$7∗$ + 7 = 7! + 7 = 1*2*3*4*5*6*7 + 7 = a Multiple of 7 => NOT a Prime Number

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