Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Combinations Help

  1. Aug 27, 2016 #1
    Ok, so the problem is:

    There are 5 letters, ABCDE. How many unique combinations of 3 letters can be made if letter A must be part of the combinations?

    The solution is that because A must be one of the letters, that only leaves two other letters which you can manipulate. So the answer is (4)(3)/(2!) = 12/2 = 6 possible unique combinations

    But my confusion arises from the explanation as to WHY the answer is calculated this way.

    For example, if the condition that A must be part of the combination, then I know the three letter combination has 5*4*3/(3!) = 10 unique combinations.

    But when I consider that A must be part of the solution, I think like this:
    1) There are 3 boxes, and one of them must be A, meaning it can only be one possible choice.
    2) The rest of the two boxes can be the other 4 choices.
    3) So my permutations are 1 * 4 *3 = 12
    4) Unique combinations are 12/(3!) = 2

    When I write out the solution, I know that my answer of 2 is wrong, and the answer is 6 is correct, but I cannot understand the reasoning behind calculating 6 the way it's done.

    I can slightly reconcile the reasoning if I think like this:
    1) I need an A in the answer, and that can be made by choosing the option A only, which is 1 option, and the unique combination of that is 1/1! = 1
    2) The rest of the two boxes can be the other 4 letters, so the unique combination of the last two boxes are 4*3/2!
    3) The total unique combination is Ways to Choose A * Ways to Choose Other Two Letters = 1/(1!) * 4*3/(2!) = 6

    But still, I'm a bit confused :(
     
  2. jcsd
  3. Aug 27, 2016 #2

    fresh_42

    Staff: Mentor

    I think you make it too complicated. Since A is fixed, you are left with a choice of 2 out of 4. This is ##\binom{4}{2} = 6##.
     
  4. Aug 27, 2016 #3
    Haha, maybe I am. But still, I'd like to know once and for all. This has been nagging for the longest time.

    So, in short, my thinking is that I can choose A to be in the first slot, which is 1 option, and the other two are going to be 4 and 3 options respectively.

    1*4*3 = 12 permutations

    But 3 slots means there are 3! permutations that are all the same combination, so 12/3! = 12/6 = 2 combinations, which is wrong. But I don't understand WHY it's wrong :(
     
  5. Aug 27, 2016 #4

    fresh_42

    Staff: Mentor

    So far so good. 12 possibilities. But here you've counted (BC) and (CB) as two. And all other pairs as well. This means you double counted and the solution is 12 / 2 = 6.

    By calculating with three arbitrary slots, A isn't fixed anymore. In this case you get ##\binom{5}{3} = 10## possibilities, of which ##\binom{4}{3}= 4## do not contain A. And ##10-4=6## again.
    However, I didn't really understand your calculation here.
     
  6. Aug 27, 2016 #5
    Yeah, this part is what trips me up because if we consider that A doesn't have to be part of the answer, then the number of unique 3 combinations is 5*4*3/3! = 10. We divide by 3! because a set of three items can be permutated in these many ways.

    But now if we consider A must be part of the answer, then the combination then becomes 4*3/2!
    I'm not sure why it's not 1*4*3/(3!) instead of 4*3/2!

    But why do I divide by 2! ?
    Aren't I choosing 3 slots because A must be part of the slots, so I should divide by 3! as that is how many ways a set of three items can be permutated?

    Can you explain what 4C3 means? I'm not sure why you did that.
     
  7. Aug 27, 2016 #6

    fresh_42

    Staff: Mentor

    Consider ##n## labeled letters ##A,B,C, \dots ## and you pick ##k## many out of them. As in our case here, we do not consider the ordering of the chosen letters. Therefore we simply have a subset with ##k## elements from a set with ##n## elements.

    The question now is: How many different subsets of this kind can be chosen?
    And the answer is ##\binom{n}{k} = \binom{n}{n-k} = \frac{n!}{k! \, (n-k)!} = \frac{n \cdot (n-1) \cdot {\dots} \cdot (k+1)}{1 \cdot 2 \cdot {\dots} \cdot (n-k)}## many different.

    So either you calculate ##\binom{4}{2}=6## for two letters (for the remaining slots) out of four letters left (without letter ##A##),
    or you calculate ##\binom{5}{3}=10## possibilities to chose three letters out of five minus ##\binom{4}{3}=4## possibilities to chose three letters out of four, i.e. those without letter ##A##.

    Because you have only two slots available, four letters for two slots. One is already occupied by letter ##A##.

    If you want to calculate with all three slots, then you have (three out of five) ##\binom{5}{3}=\frac{5\cdot 4}{2!}=\binom{5}{2}=\frac{5\cdot 4 \cdot 3}{3!}=10## minus the (three out of four) ##\binom{4}{3}= \frac{4}{1!}=\binom{4}{1}=\frac{4 \cdot 3 \cdot 2 }{1 \cdot 2 \cdot 3}=4## for which ##A## isn't in the selection.

    You simply must not mix the two different approaches.
     
  8. Aug 28, 2016 #7
    I don't get why A isn't counted as a slot, though.
    A is still a choice, as are the other two slots. So we have 3! ways to combine any 3 items.

    Ok, this method makes sense to me.
     
  9. Aug 28, 2016 #8

    fresh_42

    Staff: Mentor

    We don't consider the order of the selections. So we may pretend as if ##A## is chosen first. The results are the same as if ##A## is chosen second or third. However, one of these has to happen for ##A## being in the final selection. And only one of these! You could as well start with:
    Case 1: ##A## is selected first. Then ...
    Case 2: ##A## is selected second. Then ...
    Case 3: ##A## is selected third. Then ...
     
  10. Aug 28, 2016 #9
    It's just not clicking for me :(

    I'm sorry!

    What if I did it this way:
    1) Find all possible unique combinations in which the restricted letter can fill up 3 spaces
    2) Multiply (1) by the possible unique combinations that the rest of the letters can fill up 2 spaces

    And (2) would be my answer.
     
  11. Aug 28, 2016 #10

    fresh_42

    Staff: Mentor

    Try the cases.
    If A is chosen first, then there are 4 left for the second choice (say B) and 3 left for the third choice (say C). This makes 7 but we double counted 1 (BC and CB). So 6 possibilities are left.
    The same goes with the other cases. One of theses cases has to happen, and only one.

    I don't get this. Either you calculate with all letters, or with one fixed and then with a slot less. You cannot do both at the same time. If you restrict the number of letters (5 → 4) but allow all (3) slots it confuses me, too.
     
  12. Aug 29, 2016 #11
    Hmm, ok.

    But why do we lose a slot and divide by 2! instead of saying there are 3 slots, but it so happens one slot must be A? So there are still 3 selections which can be arranged in 3! ways, right?

    So then why don't we divide by 3! ?
     
  13. Aug 29, 2016 #12

    fresh_42

    Staff: Mentor

    Perhaps it becomes clearer if you proceed step by step and use an unambiguous language.
    I really don't know, what exactly you are talking about.

    E.g.:
    Slots aren't lost. It's always three of them. But for exactly one of them there is no choice. Now, whether you either say "that makes 3! / 3" for all or "2!" for the rest doesn't make a difference.
     
  14. Aug 29, 2016 #13
    Ok, my confusion is what to divide by.

    Situation A:
    Just considering all the ways I can make 3 permutations from 5 choices, that is 3 available slots, where the first slot has 5 choices, the second slot has 4 choices, and the third slot has 3 choices. So that is 5*4*3

    But to get the unique combinations, I have to remove all duplicates. I have to divide by 3! because for any set of 3 items, they can be combined in 3! ways.
    So I get 5*4*3/3!

    Situation B:
    But, if I say A must be in one of the slots, then that leaves only 1 choice for the first slot, 4 choices for the second slot, and 3 choices for the third slot.
    So the permutations are 1*4*3

    But, to get the unique combinations, I have to remove all duplicates. I have to divide by 3! because for any set of 3 items, they can be combined in 3! ways.
    So I get 1*4*3/3!
    (But the real answer is 1*4*3/2!)

    Now, if I write out the answer, I know it doesn't come out to be correct. But I don't understand why the reasoning I used in situation A doesn't apply for B.
     
  15. Aug 29, 2016 #14

    fresh_42

    Staff: Mentor

    Correct. This gives you 10 possibilities as expected, because you didn't impose any conditions (on A).

    Correct until here.

    And this is wrong. You said above
    which leaves you only two slots available for the next two(!) choices where you can have duplicates.

    ... because you have to divide by 2! possible duplicates.
     
  16. Aug 30, 2016 #15
    Hmm, ok.

    But if dividing by 2! means that I am selecting only 2 slots, doesn't that mean the first slot is fixed? As in A MUST be in the first slot?

    Like I can get ABC, ABD, ABE
    But not BAC, CAB, EAB (since A is not in the first slot)

    What if A is in the second or third slot?
     
  17. Aug 30, 2016 #16

    fresh_42

    Staff: Mentor

    .... plus ACD, ACE and ADE and thus getting six in total.
    Then you will get (for A selected second) BAC, BAD, BAE, CAD, CAE, and DAE and so on ...
    This is the solution which distinguishes between the three cases I mentioned earlier.

    The information ("A has to be selected") is turned into knowledge.

    When done beforehand, we count the number of remaining possibilities.
    This means, A is (knowingly) in exactly one slot. Cases 1,2,3.
    Here the information is applied prior to the counting.

    The same is done with the one step calculation: 4 (allowed) letters in 2 (remaining) slots, i.e. ##\binom{4}{2} = 6## possibilities.
    The information is applied first.

    The other possibility is to apply the information afterwards, which means we count all possibilities (##5 \cdot 4 \cdot 3 = 60##), forget about the order (division by ##3! = 6##) and subtract all possibilities, that do not contain A (minus ##\binom{4}{3}=4##).

    Your argumentation is somehow changing horses in the middle of the race. You start with the information beforehand, forget it in the middle and apply it another time at the end.
     
  18. Aug 30, 2016 #17
    Ok, so given that A must be in a slot, it could be A is in the first, second, or third slot.

    For A in first slot: 1*4*3/2!
    For A in second slot: 4*1*3/2!
    For A in third slot: 4*3*1/2!

    How do you know that each of these situations has the exact same combination as the other situations?
     
  19. Aug 30, 2016 #18

    fresh_42

    Staff: Mentor

    It hasn't. ABC and BAC are different. But they cannot occur both. The total number of combinations of the kind A** is equal to the number of combinations of the kind *A* which is also the same number of combinations of the kind **A. But there cannot be a mixture of these combinations. Each individual pattern has six possibilities. You can know it's the same number of combinations, because we don't have numbered slots. We cannot distinguish between *A* and A** other than "A is selected second" and "A is selected first".
     
  20. Aug 31, 2016 #19
    Ok, I think it's starting to make sense.

    There's also something I thought of last night. So I mistakenly assumed that the factor you divide by (x!), that the x represents the total number of slots. But actually, if x! should represent the number of different permutations, then that means each slot must be able to choose from the same pool of options as the other slots to create a permutation using those same options. So that means x should represent the number of slots that can choose completely the same pool of options as the next slot, not the total number of slots.

    So since one of the slots must only be A, it cannot choose from the pool of options that slot 2 or 3 can use.

    So then, if one slot can only be A, then the possible combination of that is one choice (A), over 1 slot that can choose that same pool of options (only A). So it's 1/1!

    And the other two slots can choose from the same pools of option (B C D E), so that would be 4*3/2!

    So in total, it is 1/1! * 4*3/2!

    That must be it :(
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Combinations Help
  1. Help with Combinations (Replies: 4)

  2. Combination of cards (Replies: 16)

  3. Booze combinations (Replies: 12)

Loading...