I did like this
D, K eat chocolate + D, K eat oatmills

For D, K eat chocolate,
DK are already selected, so 4 people left, also 2 chocolate left
so, 4C2,
ie, (Choc, Oatmill) ---> ({D, K, R , T }, {M, N })

For DK eat oatmill,
DK are already selected, so 4 people left for 1 oatmill , when 1 oatmill person is selected, the rest go to chocolate,
so, 4C1 * 3C3
ie, (Choc, Oatmill) ---> ({M, N, R}, {D, K, T})

but the answer i get is 10


where am i wrong please help

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

Case 1: 3 Chocolate and 3 Oatmeal
K has 6 options.
No matter what K chooses, D has 3 options.
The third to choose has 4 options.
The fourth to choose has 3 options.
The fifth to choose has 2 options.
The sixth to choose has 1 option.

6*3*4*3*2*1 / 3*2*1*3*2*1 = 3*4 = 12

Case 2A: 4 Chocolate and 2 Oatmeal; K chooses Chocolate and D chooses Oatmeal
K has 4 options.
D has 2 options.
The third to choose has 4 options.
The fourth to choose has 3 options.
The fifth to choose has 2 options.
The sixth to choose has 1 option.

4*2*4*3*2*1 / 4*3*2*1*2*1 = 4

Case 2B: 4 Chocolate and 2 Oatmeal; K chooses Oatmeal and D chooses Chocolate
K has 2 options.
D has 4 options.
The third to choose has 4 options.
The fourth to choose has 3 options.
The fifth to choose has 2 options.
The sixth to choose has 1 option.

Same as 2A = 4

Total disallowed options is 12+4+4 = 20.

30-20 = 15

Answer choice D.