Combinations

Probability Level 1

If the order doesn't matter, in how many ways can the first 100 natural numbers be grouped if we take 2 numbers in one group?

For example, if we have the set of numbers $(a,b,c,d)$ , then they can be arranged in the following manner:

$\{ ab , bc , cd , ac , bd , ad \} ,$

so, there are 6 ways in which these 4 numbers can be paired in pairs of 2.

Bonus: What could be the general formula for this type of scenario?

Grand Bonus: Find the sum of all the combinations.

The answer is 4950.

2 solutions

Zach Abueg
Jan 26, 2017

Related Wiki: Combinations

This is a simple combinations problem. Our set is made up of $\displaystyle 100$ elements, and we want to know how many different combinations we can make from choosing $\displaystyle 2$ of them, given that order does not matter (in other words, $\displaystyle ABC, ACB, BCA, BAC, CAB,$ and $\displaystyle CBA$ are all the same combination).

The number of combinations of $k$ objects chosen from $n$ objects is denoted by ${n} \choose {k}$ $\displaystyle = \frac {n!}{k!(n - k)!}$ .

$\displaystyle \frac {100!}{2!(100 - 2)!} = \frac {100!}{2 • 98!} = \frac {100 \times 99}{2} = 4950$

There are $\displaystyle 4950$ ways to choose $\displaystyle 2$ objects from $\displaystyle 100$ .

Conor Donovan
Jan 26, 2017

$100 \choose 2$ = 4950

Combinations

The answer is 4950.

2 solutions

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