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? 
 A: This is only $\binom{25}{5}$. Ok?
A: 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.
A: 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 $n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1.$
