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Thread: Two theoretical Probability questions - Probability 101

  1. #1
    Actuary.com - Level I Poster
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    Jul 2009

    Two theoretical Probability questions - Probability 101

    As I study for the P exam, my mind always gets distracted by things that my book either doesn't address or won't address until later (which I don't have the patience to find out, because it will bug me so much!) Here are two questions that came up.

    1) We know that when counting Permutations, and some objects are identical in the set, we simply do n!/(r1! * r2! ...) where n is the number of objects, and r1 objects are identical to each other, r2 items are identical to each other, and so on.

    Using Combinatorics, we know that (n choose r) gives us the number of possible combinations of n objects taken r at a time (when the order of selection is irrelevant).

    Now, what if some of the objects from the original group of n where in fact identical? For example, We have want to work with the word PEPPER, and we want to determine the number of possible combinations when we take 4 items at a time. It seems a little different. My intuition tells me that we just divide by 3! and 2! ...? But my intuition is normally wrong. Can someone generalize this using n's and r's? :wacko:

    2) We know that the notation (n choose n1, n2, ..., nr) represents the number of possible divisions of n distinct objects into r distinct groups, of respective sizes n1, n2... and the number of items in all of the groups sum to n. So what determines if one group is "indistinguishable" from another? For example, 10 children are going to be divided into an A and B team. How many different divisions are there? It's simply (10 choose 5, 5) = 10!/(5! * 5!). But if we just say, 10 children are going to divide themselves into two teams of 5. Now the problem is different, and the answer is [10!/(5! * 5!)]/2!

    Why is the second one different? Is it because we labeled the groups in the first example? What if we didn't label the groups, and just said that they were going to divide themselves into a group of 6 and a group of 4. Is the fact that these two groups are different in number by definition make them different? I.e., the "6" group and the "4" group. If that's the case, then the only time the second example's thought process comes into play is when 1) we don't "label" the groups beforehand and 2) ALL the groups are the exact same size, i.e., n1 = n2 = n3 = ... = nr? and 2) :shocked: And if THAT'S the case, I'm sure there's alot of linear algebra that could be done to figure out just how many times this is possible.. for example, from 10 items, you can do this with groups of 2 and 5, but not 3's because there is no linear combination of 3's using integers that can sum to 10? But don't ask me, I barely passed linear algebra.

    Thanks for the input on all or any parts of this post... It's been bothering me for a while!

  2. #2
    Actuary.com - Level I Poster
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    Aug 2009
    To sort out the "n!/(r1! * r2! ...)" business, think about the number of ways to spell banana. Well, the number of ways to arrange a word with unique letters is (# of letters)!. Then, you have to divide through by the repeats (here, if you switched the two n's, the word banana would look the same)

    In other words:

    Normal Word: 6!

    Howevever, we have nondistinct letters (a and n). We have to count the number of permutations that give the same word. You could mix the a's around 3! (6) times and the n's 2! (2) times. All together, that's 3!*2! that we need for the "correction factor" in the denominator.

    So, 6!/(2!3!) is the number of ways to spell banana uniquely with each of its letters.

    If you're still confused, here's the simple way to do it, get the following data:
    -Number of Total Letters (n)
    -Number of Repeated Letters (r)

    Take, for example, pepper

    Six letters, thus n=6

    We have triple p and double e, so the denominator is the number of ways to permute the p's times the number of ways to permute the e's, or 3!*2!

    Finally, divide n!/[(r1)!(r2)!(....)]. In this case, it would be 6!/(3!*2!)

    So yes, you were on the right track just tried to walk you through why that works.

    Also, chooses only apply to distinguishable objects. It's like asking yourself, "How many ways can I select two people from my high school." Each person is distinguishable.
    Last edited by rob22; August 23rd 2009 at 12:58 AM. Reason: Stupid typos :(

  3. #3
    Actuary.com - Level I Poster
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    Also n choose r = n! / (r! * (n-r)!)

    n permutation r = n!/(n-r)!

  4. #4
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    I understand what you are saying. However, what I meant was, what if we are wanting to count the possible number of combinations of PEPPER when we take 4 letters at a time? I'm not trying to make an actual word, so sorry for the confusion. For example, if it were just ABCDEF, the simple answer is (6 choose 4). However with PEPPER, we have 3 repeated Ps, and 2 repeated E's. So now what happens? We could select P(1)-E(1)-P(2)-R, and then R-P(2)-E(2)-P(3), however these "groups" are indistinguishable (both P-P-E-R). How would you go about counting these?

  5. #5
    Actuary.com - Level I Poster
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    Ah ok, now I understand what you are asking. First of all, I would say that with a simple word such as pepper, you are better off with a list than a formula probably. You could sort this list by first writing the ones with 3 P's, then 2 P's, then 1 P (no group of 4 can have no p obviously...). Anyways, you probably want a more in depth method than that if you were given some crazy string of letters like fjiowejfoewijfeiowjfioewjfiowjfioewjalfioorqwiojds . Well, I'll do my best to help.

    First of all, let me move to a related, but different example. Imagine you walk into a wonderful donut (or bagel, whatever) store. There are (just to make the numbers easy) 5 different types of donuts, with 20 of each ready to eat. Consider this question:

    How different combinations of donuts can you pick when buying a dozen donuts, if each type of donut (glazed, strawberry, chocolate, vanilla, sprinkles) is identical.

    For example, one type combination would be GGGGSSCCCCCV. (no sprinkles)

    In probability, this boils down to what is often called a "bars and x's" problem.

    Imagine the following x's: xxxxxxxxxxxx (there are twelve of them)

    Now I am going to insert 4 bars between these x's:


    This is a very convenient notation, because this says to me: 3 Glazed, 2 Strawberry, 4 Chocolate, 2 Vanilla, 1 Sprinkles.

    Another one, showing no sprinkles, and one more vanilla, would be: xxx|xx|xxxx|xxx| Notice how there are still 4 bars.

    We will always have (the number of donut varities - 1) bars, in this case, we had 5 donut varities, so 4 bars to make it into 5 groups.

    To organize some notation, let n=number of donuts=12, and r=number of varities=5, and remember that r-1=number of bars.

    Now, you have 16 total objects (12 x's and 4 bars, or n x's and r-1 bar's, so n+r-1 objects). Think about it this way, you have 16 blanks

    _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

    And you need to choose where to put the 4 bars. Once you choose the placement of the 4 bars, the x's will fill the remaining sllots. Note: bars can be next to each other (it means there will be none of a type of donut)

    So, once again, you have n+r-1 objects and you need to choose where to put the r-1 bars.

    Thus, the formula we has is n+r-1 choose r-1. This is also equivalent to n+r-1 choose n. Using n=12 and r=5, we get 1820 ways.

    But wait, I still didn't answer your question. Let me connect the two. In this problem, there was no shortage of any type of donuts. For example, if I got all glazed, there were still extra's left over. If there had only been 11, the story would have been a little different. In that case, you'd have to pretend like there was enough of everything, and then subtract the cases that didn't make sense. For example, with pepper, you can't choose all p's for a 4-letter group. You will run out. However, this is a similar problem which does work if you have more than enough of each type. This is the typical way to deal with a repetition allowed, order doesn't matter set. However, you have to make an assumption you did not make. I do not yet know of a way to deal with a word like pepper that you could crank out with an easy formula (but perhaps it exists). I hope this helps somewhat though. Also, I don't see them giving a ridiculous word on Exam P.

    I hope this helps.
    Last edited by rob22; August 25th 2009 at 12:22 AM.

  6. #6
    Actuary.com - Level I Poster
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    ''', thanks for the input! I'll take it all in sometime this week, have to get up early tomorrow .... Thanks again!

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