Number of possible combinations

  • Thread starter kidkook
  • Start date
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.


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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.
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.


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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##.
Excellent. This is exactly what i was looking for. A very well documented explaination. Many thanks.

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