Combinations

gmatchallenge

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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.

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 : 9X1X8X7/2! = 252. 9X1 shows same selection for two positions where originally 9 options were available. 8x7 shows the possibilities for last 2 slots which have to be populated by different varieties. 2! accounts for double repetition among 8x7.

(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 9x1x8x1. 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 8x7 only and not 9x8x7 i.e 3! - Why ?

(4). three are the same and the fourth different
A. 9x1x1x8 = 72 (Again, why are there no repetitions between repeated and non-repeated elements)

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

Will appreciate your input.
 
I'm not sure exactly what your "slot method" is, but it may need to be refined or extended a bit. Taken literally, "slots" (I think of them as "blanks", like _ _ _ _) are only directly relevant to permutations, not combinations, but they can be applied in a variety of ways. You do appear to be applying it correctly for the first two, but you are not making all the corrections needed for the others. I wouldn't use permutations, myself, but just brerak down the process of selecting using combinations.

Let's take (2): We have 9 types to choose from, and we want 2 the same, one different. I would say it this way: since order doesn't matter, we can first choose a type of which we will pick two, and then take them (no choice needed); that can be done in C(9,1) = 9 ways. Then we need to pick two of the remaining 8 types, which can be done in C(8,2) = 8*7/2 = 28 ways. This yields your answer; but I have done it using combinations rather than permutations.

Now (3): We want 2 each of 2 types. So we first pick two types, in C(9,2) ways. Then we take 2 of each type (no choice needed!). We're done; the answer is 9!/(2!7!) = 9*8/2 = 36. In your method, you neglected the fact that it doesn't matter which type comes first. You will have called AABB and BBAA two different selections.

Give (4) another try, and see if you can get it right. If you must use the "slot method", think very carefully about how to adjust for repetitions. Maybe tell us what you were taught about how to adjust, and especially how to recognize when you need to.
 
Dr Peterson,
I am a bit confused here. Let's just look at the 1st question for now. It seems as your answers are assuming that there are exactly 9 biscuits in the box. Are you assuming this? If yes, why is that the correct? I am picturing this large box of biscuits having nine varieties but having many more than 9 biscuits in total. Shouldn't that cause the problem to not have an answer? What am I missing!
 
Dr Peterson,
I am a bit confused here. Let's just look at the 1st question for now. It seems as your answers are assuming that there are exactly 9 biscuits in the box. Are you assuming this? If yes, why is that the correct? I am picturing this large box of biscuits having nine varieties but having many more than 9 biscuits in total. Shouldn't that cause the problem to not have an answer? What am I missing!
Jomo

Dr. P is talking about types, not cookies.

On question 1, we are choosing 4 biscuits, each of a different type. So the answer is

[MATH]\dbinom{9 \text { types}}{4 \text { types}} = \dfrac{9 * 8 * 7 * 6}{4 * 3 * 2} = 126.[/MATH]
On question 2, we are choosing 4 biscuits, 2 of 1 type and 1 of two different types.

[MATH]\dbinom{9 \text { types}}{1 \text { type}} * \dbinom{8 \text { other types}}{2 \text { types}} = 9 * \dfrac{8 * 7}{2} = 252.[/MATH]
And so on.

Essentially, we are assuming biscuits of the same type are indistinguishable, and that the number of ways means number of distinguishable ways. The question is a bit fuzzy in that sense.
 
Dr Peterson,
I am a bit confused here. Let's just look at the 1st question for now. It seems as your answers are assuming that there are exactly 9 biscuits in the box. Are you assuming this? If yes, why is that the correct? I am picturing this large box of biscuits having nine varieties but having many more than 9 biscuits in total. Shouldn't that cause the problem to not have an answer? What am I missing!

No, there are 9 KINDS ("varieties") of biscuits, with enough of each kind in the box that there is no limit on how many of each we can use. But each biscuit of a kind is indistinguishable from the others.

That is, the problem is equivalent to asking about words we can make using only the letters A, B, C, D, E, F, G, H, I, J, under various conditions.

So "all four are different" means they are all different KINDS; we must use four different letters.

If there were only 9 biscuits, then the subsequent problems would be nonsense -- you couldn't have two of the same biscuit! But you can have two of the same KIND.

I'm not alone in having said repeatedly that combinatorics problems have to be read very carefully, and are easy to misread (or misstate, so that they become ambiguous). And you have to be familiar with some conventions within this genre.
 
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.
(2). two are the same and the others different.
3). two each of two varieties are selected.
(4). three are the same and the fourth different
1) \(\displaystyle \mathcal{C}^9_4\)

2) \(\displaystyle \mathcal{C}^9_1\cdot \mathcal{C}^8_2\)

3) \(\displaystyle \mathcal{C}^9_2\)

4) \(\displaystyle \mathcal{C}^9_1\cdot \mathcal{C}^8_1\)

Now gmatchallenge you have the answers, you need to explain why they each work. Please do.
 
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