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Number of possible combinations

  1. May 14, 2014 #1
    I have 16 items and i need to know the number of possible combination for all 16 items. I know the possible number of combination for all 16 items is 1 and the possible number of combinations for 15 of the 16 items is 16 using the formula below:

    [tex] \frac{n!}{r!(n-r)!}.[/tex]

    How can i adaprt this formula to calculate for the number of selection from 1 to 16 inclusive without repeating the calculation 16 times.

    So n will always be 16 and r will be 1 to 16 inclusive.
  2. jcsd
  3. May 14, 2014 #2


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    Your wording is a little confusing. Is this the correct interpretation:
    You want to know the total number of ways to select at least one item from the 16.

    If it is, think about this: you should know the total number of subsets (empty, proper, with the original set) for a set with 16 items. Your question involves finding not the total number but how many there are not counting the empty set.
  4. May 14, 2014 #3
    Sorry for the confusion. What i have is a dataset with 16 different values. What i'm trying to achieve is to determine how many combination are available without repetition/duplication of a value. So i know the answer is 65536. I achieved this by using the above formula 16 times and substituting the value of r from 1 to 16 etc...and adding the number together.

    Number of combinations Value of r
    16 1
    120 2
    560 3
    1820 4
    4368 5
    8008 6
    11440 7
    12870 8
    11440 9
    8008 10
    4368 11
    1820 12
    560 13
    120 14
    16 15
    1 16

    What i would like to know is...can the formula be adapted to so i don't need to repeat the calculation 16 times. My dataset is likely to increase significantly in the future so it will not be practical to use this method.
  5. May 14, 2014 #4


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    Note that ##\frac{n!}{r!(n-r)!}## is called the binomial coefficient and is often denoted
    $${n \choose r} = \frac{n!}{r!(n-r)!}$$
    The reason it is called the binomial coefficient is the following property:
    $$(x+y)^n = \sum_{r=0}^{n}{n \choose r}x^r y ^{n-r}$$
    If we plug in ##x=y=1## then this reduces to
    $$2^n = \sum_{r=0}^{n}{n \choose r} = 1 + \sum_{r=1}^{n} {n \choose r}$$
    where the second equality follows because ##{n \choose 0} = 1##. Therefore,
    $$\sum_{r=1}^{n} {n \choose r} = 2^n - 1$$
    which is consistent with your calculation since ##2^{16} - 1 = 65535##.
  6. May 14, 2014 #5
    Excellent. This is exactly what i was looking for. A very well documented explaination. Many thanks.
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