In how many ways can X distribute 20 chocolates among A,B,C and D

(I) If each receives atleast one

(II) If X may chose to give no chocolates to whom he dislikes on that particular day.

I am really not sure about that one...on test day I would have chosen A

on B, the verb "may" : we don't know whether someone will receive chocolates or not...

Can someone tell me if the number of choices in statement A would be 20*19*18*17 ?

thanks nocilis , i know i don't need the answer but actually i am interested in the way to find the answer with statement 1 if it was required

You do need the answer Antmavel. Nocilis was trying to tell you that this is NOT a DS question. This is TWO PS questions (without given choices). In other words there's no (D) choice, or (A) choice, for that matter.

I thought that (I) was a permutation problem because order is not significant. i go 20P3.

What does everyone think?

why would it be 19c3? when the it seems that ALL THE chocolates have to be distributed amongst them

The first is a simple problem solving question, so SUFFICIENT.

In the second, X may choose, X may not choose. Even if this was not the dilemma, "to whom he dislikes" can be one or more people. Thus we do not know how many people would receive the chocolates, and how many don't. Thus, INSUFFICIENT.

I just tried "solving" the first question. My answer is different from yours.

For distributing 20 chocs amongst 4 people where each receives at least 1, distribute 1 choc each to everyone. The first 4 chocs can be distributed in 20*19*18*17 ways. Thus now everyone has 1 choc each, and 16 chocs remain. Now 16 chocs can be distributed amongst 4 people in 4^16 ways. This of course if the chocs are distinct. Total ways = 20*19*18*17*(4^16)

If not distinct, first dictribution of 4 similar chocs to 4 ppl can be done in one way only. After that 16 chocs can be distributed to 4 people in P(16, 4) ways. Thus total number of ways = P(16,4)

I tried reducing the scale and testing if it really works.

Assume 5 chocs and 3 people. My answer is P(3,2) = 6 for similar chocs

See the distribution for similar chocs (the number represents the total chocs - doesn't make any difference because all chocs are similar)

1 1 3
1 3 1
3 1 1
2 2 1
2 1 2
1 2 2

For dissimilar, assume 3 chocs (A, B, C) and 2 ppl. Answer should be 3*2*(2^1) = 12 ways.
_________________

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why is this not a DS problem guys?

(1) Therefore total outcome is 16^4*P(20,4).

(2) 20^4

Are the above answers correct?. Does it matter if the chocolates are the same or different?. What if 4 different kinds of chocolates that add up to 20? What is the OA?.

Wow that was a great thread. So according to the formula: C(n+r-1,r)
For question 2, n=4, r=20 so the total outcome = C(23,20)=C(23,3) right?

Now how do we understand this formula?

We need to divide the 20 chocolates into four shares, so we need to insert three divisors. In other words, we would have 23 empty slots, and then we pick three slots for the divisors and all the other slots would be the chocolates. So the total number of ways would be C(23,3). Got you, thanks!