Sweet Distribution

Question

During the festival of Diwali, as a popular custom, people distribute sweets among their friends and neighbour. Many a times, because of growing number of sweet boxes to distribute, people generally gift the boxes they recieve to someother person :). This is benificial in two ways, first, economically, secondly this practice discourages the stocking of sweet boxes which may get rotten over the time because of non-consumption. Now let us assume there are $n$ families who are celebrating diwali and are practicing the above mentioned custom of distributing sweet boxes. What minimum amount of sweet boxes (in total) will suffice the distribution among them, such that each of them gift sweet boxes to every other family, and are left with atleast one sweet box $?$. HAPPY DIWALI !!

Comments

umm sry....i guess your and my logic is same. I think my counting was not correct instead of doing addition , i did multiplication accidentally, now i realize my mistake. was your logic to give n-1 boxes to 1st family and then n-2 to 2nd oneand so on. First family will distribute n-1 boxes to n-1 family and then second family with n-2 boxes + 1 received from 1st family will make it as n-1 again so that it can redistribute again and so on but last family should have 1 box instead of zero right else sum would become n-1 and it would left with no boxes after distributing. were you thinking the same??

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after reading digvijaysingh answer, even his approach seems to be correct. :(

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Happy Dipawali!

Ans : n

Answer 1

According to last statement every family will be with atleast 1 box. Suppose initially every family has 1 box. Since there are n families so we numbered them as f1,f2......fn.

Now f1 can send it's box to f2 and f2 can resend the same box to f1. similarly f1 can send to f3 and received back the same box. So f1 can send it's box to (n-1) families and received back the same box from each and every family. Now f2 can send it's box to (n-2) families, f3 can send it's box to (n-3) families and so on..... the last family fn will send it's box to (n-n) i.e. 0 family.

That means if every family have atleast 1 box then we can satisfy the condition of question

So minimum number of boxes = 1*n =n

I think it is either a trick question or a poorly framed question. Becoz the answer is itself in the question

Comments

bro , what do you think about the above approach that i have taken? is this could be a possible logic to solve this kind of problem?

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U are counting number of methods of distribution not the number of boxes

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ohk..thanks for the feedback

Answer 2

Answer should be ⁿC₂

It is same as handshaking problem.

Two family disributing sweets among them.

So, if one family celebrate sweet with other family, there is no need meet them again.

Total sharing will be ⁿC₂ =n(n-1)/2

Comments

but that won't be minimum.