\(n∗\) 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
.
.
.
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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