In Combinations and Permutations is?

5 items choose 3 order not important repetition is allowed

and

7 choose 3 order not important repetition not allowed

Both 35 because you are replacing the repetition in the first example with 2 more items to choose from in the second example?

2 Answers

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  • 6 months ago

    You are correct -- the numbers do work out the same. I will show how they are related.

    For the first case, let's imagine we had 5 flavors of ice cream (Banana, Chocolate, Lemon, Strawberry and Vanilla). You are picking 3 scoops and *are* allowed to repeat flavors, if you like.

    Let's line up their containers next to each other as follows:

    B C L S V

    Next imagine you wanted to create a routine for an automated scooper telling it to pick 3 flavors. The scooper starts on the first flavor and ends on the last flavor.

    The two instructions you can give the machine are:

    - Scoop the current flavor (◯)

    - Move to the right (➔)

    So if you wanted 3 scoops of chocolate it would be:

    ➔◯◯◯➔➔➔

    If you wanted 2 scoops of Lemon and 1 scoop of Vanilla it would be:

    ➔➔◯◯➔➔◯

    Or if you wanted 1 scoop of Chocolate, Strawberry and Vanilla it would be:

    ➔◯➔➔◯➔◯

    Every combination with repetition allowed would correspond to one of these patterns. So if we can count these patterns, we can count the ways to pick 3 scoops of ice cream, repetition allowed.

    Notice that there are always 3 circles (3 scoops of ice cream) and 4 arrows (we need to move 4 times to go from the 1st to 5th container).

    In this case, therefore we have 7 symbols and we are picking the position for 3 times we scoop. This is the same as the second expression where you have 7 items (symbols in the instruction) and you are picking which 3 are the "scoop" command.

    To generalize this, if we have n items to pick from and we want k of them (allowing for repetition), that's the same as a string of n + k - 1 items and picking the k to select.

    In your case, with 5 ice cream flavors and 3 scoops, this does turn into (5 + 3 - 1) = 7 items where we are picking 3.

    C(7,3) = 35

    Notice it's not just always add 2. It was "add 2" in this case because we were picking 3 flavors. If you were picking 2 flavors, it would be add 1. If you were picking 4 flavors, it would be add 3, etc.

    Make sense?

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  • 6 months ago

    I think you are saying, assuming 5 unique items (Such as number 1-5), how many ways can you order a sequence of 3 numbers, with repetition being allowed. The answer is 125.

    For 7 unique items (1-7) with repetition not allowed, there are 210 possible combinations for a sequence of 3.

    An explanation:

    a)

    For each "spot" in the sequence, there are 5 possible options. For a single digit, this would yield 5 possible results. For 2, this would be 5*5, for each of the first results, there are now five. (From 1 to 11,12,13,14,15)

    This makes the answer 125 (5*5*5).

    b)

    Using this same formula, but without repetition, the answer is 7*6*5 because you lose one option when you use it. This equals 210.

    • Puzzling
      Lv 7
      6 months agoReport

      You've misinterpreted the question and are calculating a *permutation* where order *does* matter. The question asks about combinations where order *doesn't* matter. The question is why having 5 items and picking 3 *with* repetition equals having 7 items and picking 3 *without* repetition.

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