A woman has seven cookies - four chocolate chip and three

We can denote each child by the first letter of his or her name.

We have 2 scenarios: 1) D and K both have chocolate chip cookies, and 2) D and K both have oatmeal cookies.

Scenario 1:

After D and K have 2 chocolate chip cookies, the woman has 2 chocolate chip (c) and 3 oatmeal (o) cookies for the remaining 4 children. She can distribute the cookies to N-R-M-T as follows:

c-c-o-o (and 2 c’s and 2 o’s can be arranged in 4!/(2!2!) = 24/(2 x 2) = 6 ways)

c-o-o-o (and 1 c and 3 o’s can be arranged in 4!/3! = 24/6 = 4 ways)

We see that there are 10 ways to distribute the cookies in this scenario.

Scenario 2:

After D and K have 2 oatmeal cookies, the woman has 4 chocolate chip (c) and 1 oatmeal (o) cookies for the remaining 4 children. She can distribute the cookies to N-R-M-T as follows:

c-c-c-c (and the 4 c’s can be arranged in only 1 way)

c-c-c-o (and 3 c’s and 1 o can be arranged in 4!/3! = 24/6 = 4 ways)

We see that there are 5 ways to distribute the cookies in this scenario.

Therefore, the woman has 10 + 5 = 15 ways to distribute the 7 cookies to 6 children.

Answer: D