Factorial question

My guess would be to count the number of 2 primes in each even factor (since none in odd). So
12 has 2 in 2*2*3
10 has 1 in 2*5
8 has 3 in 2*2*2
6 has 1 in 2*3
4 has 2 in 2*2
2 has 1 in 2*1

so sum is 10 and ans is D

can you explain your answer in a broader description. Thanks

... well 12! = 12*11*10*9*...*2*1
we are asked how many times will 2 be divided within 12!. So we can ignore all the odd terms since 2 is never a multiple of an odd number. We are left with 12,10,8,6,4,2. Now we must essentially find all the prime factors of each of these numbers noting only the number of times a 2 is found in each.

the prime factors of 12 are 2*2*3 so there are two 2s
the prime factors of 10 are 5*2 so there is only one 2
etc. as shown above.

Since all these prime factors are multiplied together you are essentially breaking down 12! into it's prime factors

so 12! = 2^10*3^(don't care)*5(don't care)*7...

Hope this helps