Question

Dolve
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How will you distribute &8
choclater in 2ta
children How many each will get?
Fach will get !
choclates in each children​

Answer 1

Answer:

I will tell you a general prescription.

Let xi denote the number of chocolates i-th child gets. Then the problem translates to finding nonnegative integral solutions xi satisfying:

x1+x2+...+x8=20 (1)

where xi s are non-negative integers. That means, we are allowing for some children to get 0 chocolates.

Now you impose that each child must get at least 1 chocolate. This amounts to saying:

ui=xi−1≥0

Subtract 8 from both sides of the equation(1) to get

u1+u2+...+u8=12 (2)

where again ui≥0 , equality holds when ui=1 , that is, each child gets 1 chocolate, and satisfies the condition.

Now you say 2 of the 8 children must get at least 2 chocolates. Let's assume those 2 lucky children are u1,u2 . Then subtract 2 from both sides of equation(2) to get

v1+v2+u3+...+u8=10 (3)

where vi=ui−1≥0 . So effectively, vi=xi−2≥0 , and we are allowing x1,x2 to have at least two chocolates.

So far so good.

Now, you need to find nonnegative integral solutions of (3). How do you proceed? You have 10 items now, to be divided among 8 children (or boxes, if I may). Imagine 7 sticks, and try to distribute them among these 10 items. Any particular arrangement of the 7 sticks and 10 chocolates is bound to give you 8 partitions. We will consider the number of items in such a partition to correspond to number of items in that box. If 2 or more sticks are placed consecutively in such an arrangement, we consider those partitions to contain 0 items. Note that having 0 items in a partition does not correspond to having 0 chocolates, because 0 items will correspond to either ui=0 or vi=0 , which is actually x=1 or x=2 respectively.

So there you have your solution! In how many ways can we arrange 7 sticks and 10 items? Noting that actually 7 sticks are identical, and so are 10 items, you have N=17!7!10! , which is nothing but 17C7 !

One final point. Note how we chose specificly x1,x2 to be the "lucky children"? Well, that choice is not unique, and can be performed in 8C2 number of ways.

So, the total number of ways the task can be performed is N=8C2×17C7 .

Hope this helps.

EDIT: As the question asks that exactly two students must get at least two chocolates, we realise that the only way that condition can be satisfied is if the rest of the chocolates are distributed to the two "lucky" children that we selected. This amounts to looking for non-negative solutions of the equation:

v1+v2=10 (4)

which can be done in 11C1 ways, by the previous logic. Hence the total no of solutions now should be: N=8C2×11C1 = 308 ways.

Step-by-step explanation:

Hope u got it