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This is a 2-part PS (without answer choices) not a DS

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.

Both these problems are repetitive combination problem:
1) # of ways 20 chocolates can be distributed to 4 ppl so that each receives atleast 1 = 19C3
2) 0 can be included, then 23C3

This is a 2-part PS (without answer choices) not a DS

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.

Altogether different answer [#permalink]
16 Mar 2005, 11:32

1

This post received KUDOS

Quote:

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.

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. _________________

Re: PS: permutation - combinations [#permalink]
17 Mar 2005, 08:31

Because it is marked as PS in the thread title.

swath20 wrote:

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.

Assuming all chocolates are different from each other:

(1) Each at least receive one means ABCD each gets one chocolate. The choices is P(20,4). Then we just need to see who gets the rest of 16 chocolates. Each chocolate faces a pool of four people to choose from, the choices is 16^4. Therefore total outcome is 16^4*P(20,4).

(2) 20^4

However it may be more reasonable to assume that each chocolate is NOT different from each other. Anybody know how to do this problem if this is the case?

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!