# Thread: Permutations and combinations Help

1. ## Permutations and combinations Help

Sorry guys one more question i need help on. Thanks a lot. I really appreciate this. The question is:

There are 6 numbers, drawn at random, between 1-49. Once a number is chosen, it no longer can be used again. Each ticket has 6 numbers.

Grand prize:all numbers match the 6 numbers drawn
2nd prize: 5 numbers match
3rd prize: 4 numbers match
4th prize: 3 numbers match

How many ways are there for selecting the numbers on a ticket that can win the grand prize?
I do the same to find out for the 2nd, 3rd, and 4th prizes.

Thank you for any help given.

2. Hi.

Originally Posted by ajmaal14
Sorry guys one more question
You don't need to feel sorry for that

Originally Posted by ajmaal14
The question is:

There are 6 numbers, drawn at random, between 1-49. Once a number is chosen, it no longer can be used again. Each ticket has 6 numbers.

Grand prize:all numbers match the 6 numbers drawn
2nd prize: 5 numbers match
3rd prize: 4 numbers match
4th prize: 3 numbers match

How many ways are there for selecting the numbers on a ticket that can win the grand prize?
I do the same to find out for the 2nd, 3rd, and 4th prizes.
Well, at first you draw one of the numbers - there are 49 different balls;
you don't use this number again, so there are (49-1)=48 balls left
2. draw: you draw one number out of 48 => 48 possibilities for that
You do not use that number again, so there are 47 numbers left
3. draw: you draw one out of 47 (47 possibilities)
4. draw: one out of 46
5. draw: one out of 45
6. draw: one out of 44

So there are 49*48*47*46*45*44 different ways of drawing 6 numbers out of 49.

Edit: I made a mistake. Sorry
Plato gave the right answer: Because of
A lottery ticket is treated as a combination not a permutation, order makes no difference.
you need to divide by 6!, thus
$\frac{49*48*47*46*45*44}{6*5*4*3*2*1} = \begin{pmatrix} 49 \\ 6 \end{pmatrix}$
is the solution!

Yours
Rapha

3. Originally Posted by Rapha
By the way, it is
$49*48*47*46*45*44 = \begin{pmatrix} 49 \\ 6 \end{pmatrix}$
Actually it should be $\frac{49*48*47*46*45*44}{6*5*4*3*2*1} = \begin{pmatrix} 49 \\ 6 \end{pmatrix}$
A lottery ticket is treated as a combination not a permutation, order makes no difference.