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While preparing for GMAT I came across one of the questions.

I have been able to apply concepts to 4 out of 5 sub questions but even after spending a day trying to look for similar questions and solutions I am stumped.

Basically, I am unable to identify that when we use fundamental counting principle (such as in this case), how do we determine that how many permutations have we repeated so that we can compensate for over estimation.

Q. A large box of biscuits contains nine different varieties. In how many ways can four biscuits be chosen if:

(1). all four are different.

A. So this one is pretty straight forward. We have to select 4 different types of biscuits so we have 9C4 or 124 ways.

(2). two are the same and the others different.

A. This was a tough one for me and even though I got the answer correct, I am not sure if my approach was correct. Switching to slot method : 9*1*8*7/2! = 252. 9*1 shows same selection for two positions where originally 9 options were available. 8*7 shows the possibilities for last 2 slots which have to be populated by different varieties. 2! accounts for double repetition among 8*7.

(3). two each of two varieties are selected.

A. I cant seem to make logic for this one. Going by the slot method I think it should be 9*1*8*1. However, looking at the answer there appear to be some repetitions that I haven't taken out. How can there be repetitions ?

Slot 1 : 9 options

Slot 2 : Same as slot 1

Slot 3 : 8 options

Slot 4 : Same as slot 3.

If there are repetitions here then why didn't we account for them in (2) among repeated and non-repeated elements. i.e in (2) there were repetitions among 8*7 only and not 9*8*7 i.e 3! - Why ?

(4). three are the same and the fourth different

A. 9*1*1*8 = 72 (Again, why are there no repetitions between repeated and non-repeated elements)

I understand that permutations/FCP take AB and BA as 2 different outcomes and that is something that needs to be compensated for but how do I calculate the repeated outcomes that need to be compensated in case of sampling where part of it is repeated and part is different e.g Q2-4 in this case.

This question looked basic but this has challenged my understanding and concepts regarding P&C with repetitions.

Will appreciate your input.

Ahmad R.
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  • In (2) there were “repetitions among 87 only and not 98*7” because AABC is the same as AACB, but not the same as BBAC or CCBA. You could have also solved (2) by 9C3 x 3C1, because you are choosing three varieties, then choosing one of those three to have two of. In (3), AABB is the same as BBAA, so you have to divide by 2. In (3), you’re just choosing 2 varieties, so it’s 9C2. In (4), AAAB is not the same as BBBA, which is why you don’t need to divide by 2. You could solve (4) by 9C2 x 2C1, because you’re choosing 2 varieties, then choosing 1 of the 2 to have 3 of. – Joe Oct 19 '19 at 12:17
  • Presumably order of selection doesn't matter. As such, it is impossible to tell which flavor was the "first pair" and which was the "second pair" so picking them one after the other in distinct order incorrectly applies significance to the order in which you picked them. The answers are: 1) $~\binom{9}{4}~$, 2) $~9\cdot \binom{8}{2}~$, 3) $~\binom{9}{2}~$, 4) $~9\cdot 8$. – JMoravitz Oct 19 '19 at 12:18
  • Your error in part 3) is effectively the same common error made in the problem of calculating the probability of getting a 2-pair in a game of 5card poker. See here. – JMoravitz Oct 19 '19 at 12:20
  • Follow-Up Question : @joe In (2) will 987 take into consideration arrangements such as ABCA, ACBA, BACA, CABA since they can be any order ? I mean they dont have to be in the form AABC or AACb right ? I get that only half of the permutations will count. But the permutations in (2) will have 2 letters of one type and 2 letters of different types in every possible order ? Is my understanding correct so far ? – Ahmad R. Oct 19 '19 at 17:25
  • @Joe Sorry I didn't understand the last sentence. 9C2 means that we are selecting 2 elements and then 2C1 means that we are selecting 1 from the 2 so shouldn't that make it a total of 2 + 1 = 3elements instead of 4 in a single selection? As you said in (2) 9C3 means we are selecting 3 and then 3C1 means we are selecting 1 from the 3 so we have 3 + 1 = 4 total elements in a single selection. – Ahmad R. Oct 19 '19 at 17:41
  • @Ahmad, I don’t follow your first comment. For your second comment, 9C2 x 2C1 is because you’re choosing two varieties from 9, then choosing which of those two varieties to have 3 of. It doesn’t matter how many elements (treats) you’re choosing. The number of “ways of choosing” is based on the varieties, not the treats. How many ways can you choose 17 treats, if you choose 6 treats of one type and 11 of another type? As long as there are at least 11 treats in each variety, the answer is 9C2 x 2C1. The “17”, “11”, and “6” have no part in the answer. – Joe Oct 19 '19 at 18:49
  • I’ll add that if it was, “How many ways can you choose 14 treats, if you choose 7 of one variety and 7 of another?”, then the answer would be 9C2. You wouldn’t multiply by 2C1, because you are only choosing 2 varieties out of 9; there’s no second choice to make, since you’re taking 7 of each variety. Whereas, if asked how many ways there are to choose 14 treats, taking 8 of one variety and 6 of another, then the answer is 9C2 x 2C1. Does that make sense? – Joe Oct 19 '19 at 20:49

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