Bunuel wrote:
A guardian with 6 wards wishes every one of them to study either Law of Medicine or Engineering. Number of ways in which he can make up his mind with regard to the education of his wards if every one of them be fit for any of the branches to study, and at least one child is to be sent in each discipline is:
(A) 120
(B) 216
(C) 329
(D) 540
(E) 729
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions Solution:Since at least one child is sent to each discipline, the number of children in the three disciplines must be distributed in one of the following ways:
4 children in one discipline, 1 child in each of the remaining disciplines
3 children in one discipline, 2 children in another discipline, 1 child in the remaining discipline
2 children in each discipline
Let’s calculate the number of possibilities in each scenario:
4 children in one discipline, 1 child in each of the remaining disciplinesLet’s first calculate the number of ways in which 4 children study law, 1 child studies medicine, and 1 child studies engineering. There are 6C4 = 6! / (4!*2!) = 3 x 5 = 15 ways to choose the children who will study law. Since there are two children left, there are two choices for the child who will study medicine. The remaining child will study engineering. Thus, there are 15 x 2 x 1 = 30 ways in which 4 children study law, 1 child studies medicine and 1 child studies engineering.
Now, let’s calculate the number of ways in which 4 children are in one discipline and 1 child is in each of the remaining disciples. We can have 4 children in one discipline and 1 child in each of the remaining disciplines in three ways:
4 children in law, 1 child in medicine, 1 child in engineering
1 child in law, 4 children in medicine, 1 child in engineering
1 child in law, 1 child in medicine, 4 children in engineering
Notice that the number of ways for each of the three scenarios above are equal, and we have already calculated that there are 30 ways for the first scenario. Thus, in total, there are 3 x 30 = 90 ways in which 4 children study one discipline and 1 child studies each of the remaining disciplines.
3 children in one discipline, 2 children in another discipline, 1 child in the remaining disciplineLet’s first calculate the number of ways in which 3 children study law, 2 children study medicine and 1 child studies engineering.
There are 6C3 = 6! / (3!*3!) = 20 ways to choose the three children to study law. Since there are 3 children left, there are 3C2 = 3 ways to choose the children to study medicine. The only remaining child will study engineering. Thus, there are 20 x 3 = 60 ways in which 3 children study law, 2 children study medicine and 1 child studies engineering.
Now, let’s calculate the number of ways in which 3 children are in one discipline, 2 children are in another discipline and 1 child is in the remaining discipline. Since there are 3! = 6 ways to arrange the digits 1-2-3, there are 6 ways to distribute the children into the disciplines such that there are 3 children in one discipline and 2 children in another discipline. For each of these distributions, there are 60 ways (since each of these distributions is identical to the case where 3 children study law, 2 children study medicine and 1 child studies engineering). Thus, there are 6 x 60 = 360 ways in which 3 children study one discipline, 2 children study another discipline and 1 child studies the remaining discipline.
2 children in each disciplineThere are 6C2 = 6! / (4!*2!) = 15 ways to choose the children who will study law. Since there are 4 children left, there are 4C2 = 4!/(2!*2!) = 6 ways to choose the children who will study medicine. The remaining two children will study engineering. Thus, there are 15 x 6 = 90 ways in which 2 children study each of the disciplines.
In total, there are 90 + 360 + 90 = 540 ways to educate the children where there is at least one child in each discipline.
Answer: D _________________
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