theBrahmaTiger wrote:
If Sandy brings her husband to the restaurant, then there are two plates that would need to be served.
There are 10 kinds of lunches available and 2 lunches would be chosen.
Hence, by combination -> 10C2 -> 45.
Now, if Sandy brings 1 child along with her husband. Then 3 lunches would have to chosen from 10 choices.
10C3 = 120.
Finally, If Sandy brings 2 kids and her husband -> 10C4 = 210.
Therefore, answer is A,C, and D.
The question is not about combination it is about permutation
Let us say that 3 types of lunch (L1, L2 and L3) are available and 2 people have are ordering
In case if they order different dishes
There are three options for first person and 2 options for the second person. So 6 orders are possible.
3p2 =6 while 3c2 =3 .
So the answers will be
10p2 = 90
10p3 = 720
10p4 = 5040
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