How many combinations of 2 can 5 make?

Answer 1

There are multiple ways to go about doing this problem.

So you have 5 items and you can choose 2 of them. Also remember that the order of the combinations do not matter (a,b is the same as b,a).

Method 1:

The way that most people would do this problem is by writing out every possible combination.

For this problem, say you have 5 letters (A, B, C, D, and E), and you want to know how many combinations you can make with 2 at a time (can't choose the same letter twice). In this method, try to make sure that your combinations are organized well. Let's write the combinations out.

With 'A' first

A B

A C

A D

A E

With 'B' first

B A (already used this one)

B C

B D

B E

With 'C' first

C A (already used)

C B (already used)

C D

C E

With 'D' first

D A (already used)

D B (already used)

D C (already used)

D E

If you count up the number of combinations that are not crossed out, you get 10 as your final answer.

Method 2:

However, in my opinion, there is a faster way to do this problem: by using the formula for combinations.

Combinations are usually notated as:

nCr

This is read as: "n choose r"

n is the number of items you have, and r is the number you are choosing, and the capital 'c' shows that you have combinations.

In your case, you have 5 items and you are choosing 2 of them at a time, so you have 5C2.

Now, on to the formula:

nCr = n!/((r!)(n-r)!)

This might seem a little strange to you, especially the exclamation points.

These exclamation points represent something called a factorial.

A factorial of a nonnegative number, n, is the product of all the positive integers less than or equal to n.

So,

1! = 1

2! = 2*1 = 2

3! = 3*2*1 = 6

4! = 4*3*2*1 = 24

...and so on.

Now that you understand factorials, let's plug in the numbers for the formula and solve.

nCr = n!/((r!)(n-r)!)

5C2 = 5!/((2!)(5-2)!)

= 5!/((2!)3!)

=(5*4*3*2*1)/(2*1*3*2*1)

= (5*4)(3*2*1)/(3*2*1)(2*1)

= (5*4)(3*2*1)/(3*2*1)(2*1)

= (5*4)/(2*1)

= 20 / 2

5C2 =10

This method may have seemed slow and tedious, but with enough practice, it will be much faster than the other method. Besides, the other method only works well with small numbers. (For big numbers always use this method, if possible).