# What are the number of ways in selecting 5 elements from 25 unique elements? The order of the selected elements are not important

What are the number of ways in selecting 5 elements from 25 unique elements? The order of the selected elements are not important?

I have 25 numbers, 1,2,3,4,5......23,24,25.

What are the number of ways a pair 5 elements be selected. The order of the selected elements are not important.

like the pair of 1,2,3,4,5 is same as 2,3,4,5,1

The order of the elements in the pair is not important.

How may number of pairs can be formed?

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Have you learned about permutations, variations and combinations? Which one applies in this case, and why? –  Pedro Tamaroff Aug 29 '12 at 22:38
This is your second combinatorics question in rapid succession. Is this homework? If it is, please use the homework tag. Also, I removed several tags from this post that don't apply. –  mixedmath Aug 29 '12 at 22:40

HINT

Let's suppose the order did matter. Then we have $25$ elements for the first choice, $24$ elements for the second choice, etc.

Of course, we're overcounting. In particular, for any choice of $5$ elements, we've overcounted it by supposing that order mattered. How many ways can we order $5$ different things? This is the number of times we counted each (order-not-mattering) choice of $5$ elements, so we should divide by the number of ways we can order $5$ different things.

Thus, our answer should be the number of choices assuming order matters divided by the number of ways we can order $5$ elements.

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So for suppose the order did matter then the number of ways would be 25*24*23*22*21 And now it should be divided by the number of ways 5 different things can be ordered which is 5! (25*24*23*22*21)/5! Is this correct? –  VIJAY AKULA Aug 29 '12 at 22:42
@VIJAYAKULA also notice that, 25*24 .. * 21 = 25!/20!. Now check out the wikipedia for binomial ceoff and combinations. –  user2468 Aug 29 '12 at 22:47
@VIJAY: Yes. You might also notice that this ends up being $25!/(5!20!)$, and this is not a fluke. I suppose that you will soon learn about combinations and permutations, and these formulas will become very familiar to you. –  mixedmath Aug 29 '12 at 22:49

This is only $\binom{25}{5}$. Ok?

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First you should ask yourself: "How many ways are there to choose five objects from 25?" You have 25 to choose from for your first object. Then you have 24 left from which to choose your second object. Then you have 23 left from which to choose your third object. Then you have 22 left from which to choose your fourth object. Then you have 21 left from which to choose your fifth object. That means there are $25\times 24 \times 23 \times 22 \times 21$ ways of choosing five objects from 25.

But wait! You said that the order doesn't matter. You need to ask yourself "How many ways could I have picked out those five objects?" This is the same as asking: "How many ways can I rearrange five objects?" You have five objects from which to choose your first, four from which to choose your second, three from which to choose your third, two from which to choose your fourth and one from which to choose your fifth. That means there are $5\times 4 \times 3 \times 2 \times 1$ ways of rearranging five objects. Putting all of this together, the answer you're looking for is:

$$\frac{25\times 24 \times 23 \times 22 \times 21}{5\times 4 \times 3 \times 2 \times 1} = 53,130 \, .$$

In general, if you want to choose $r$ objects from a pool of $n$, and the order doesn't matter, there are $n$-choose-$r$ ways of doing that:

$$^nC_r = \frac{n!}{r!(n-r)!} \,$$

where "$n$-factorial" is $n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1.$

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