# how do I calculate the different number of combinations?

theres a lottery with 6 winning numbers and the numbers range from 1-40. Would the possibe combinations be 6^40 or 40^6

### 5 Answers

- PinkgreenLv 71 month ago
The possible combination=

40C6=

40!/36!6!=

40*39*38*37*36*35/(6*5*4*3*2*1)=

10*13*19*37*6*7=

3838380

- llafferLv 71 month ago
It would be 40^6 if the numbers can be drawn more than once (replacement)

But it's not the case. Here's how the logic would work:

The first number can be any of the 40 numbers.

The second number can be any of the unused 39 numbers.

The third number can be any of the unused 38 numbers.

forth - 37 numbers

fifth - 36 numbers

sixth - 35 numbers

If we multiply these together we get:

40 * 39 * 38 * 37 * 36 * 35 = 2,763,633,600

or if you want to use factorial notation:

40! / 34! = 2,763,633,600

But this isn't the number as the order the numbers are drawn doesn't matter in lotteries. In this case, the following sets of numbers count as separate sets:

1, 2, 3, 4, 5

2, 3, 4, 5, 1

5, 4, 3, 2, 1

Since order doesn't matter we need to take out duplicates like the three sets above. To do that we divide by 6! (since there are 6 numbers)

This gives us:

2763633600 / 6!

2763633600 / 720

3,838,380 different combinations of winning numbers in this lottery.

- nbsale (Freond)Lv 61 month ago
Neither.

6^40 has no bearing on the question.

40^6 would be the number of possible sequences if repeats were allowed--you would have 40 choices for each of the six.

The correct approach works like this:

You have 40 choices for the first number.

That can't repeat, so you have only 39 for the next.

Then 38, 37, 36, 35.

Multiply those all together and you get 2,763,633,600.

But you aren't quite done yet, because you have repeats, such as 1 2 3 4 5 6 and 6 5 4 3 2 1. Those are different permutations but the same combination. Each of those are repeated 6! = 720 times, so you need to divide the above number by 720, to get the final answer, 3,838,380.

That's how you derive the well-known formula for the number of combinations of n objects, taken k at a time:

nCk = n! / [ (n-k)! k! ]

n! / (n-k)! give you the number of permutations of n things taken k at a time, and further dividing by k! eliminates the repeats and gives you the number of combinations.

These are very basic concepts in combinatorics or finite math, so if you are studying this in math, you should learn this very well. It's very important.

- What do you think of the answers? You can sign in to give your opinion on the answer.
- MichaelLv 71 month ago
This is a Combination problem

₄₀C₆ = 40! / [ (40-6)! 6! ]

₄₀C₆ = 40*39*38*37*36*35 / (6*5*4*3*2*1)

₄₀C₆ = 2,763,633,600 / 720

₄₀C₆ = 3,838,380 <––––––