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In one of the Massachusetts state lotteries, the winning number is chosen by picking six ping-pong balls from a bin containing balls labeled "1" through "36" to arrive at a sequence of six numbers between 1 and 36. Ping-pong balls are not replaced after they're chosen; that is, no number can appear twice in the sequence. How many possible outcomes are there?

Note that in this last exercise, the order in which the ping-pong balls are chosen is relevant.

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Hi Briana. You've asked five questions and haven't accepted the answers to any of them. People will be much more willing to help you out if you accept some of the answers to your previous questions. –  Chris Taylor Sep 27 '11 at 22:35
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Also, to get you started on the route to an answer: There are 36 different ways of picking the first ball of the lottery. Once you've chosen the first ball you can't choose it again, so there are only 35 ways of choosing the second ball, which means there are 36 * 35 ways of choosing the first two balls. Can you apply this reasoning to the other balls? When you've done this, don't forget that the order of the balls doesn't matter, so you should divide by the number of ways of rearranging the balls that you've chosen. –  Chris Taylor Sep 27 '11 at 22:41
    
Actually, it isn't clear that you need to divide: the order is "relevent" (I assume this is the same as relevant) –  Bill Michell Sep 27 '11 at 23:04
    
The detailed question is always below; please don't put things in the title that would apply to all questions. –  joriki Sep 28 '11 at 0:11
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@Bill, feel free to edit the question to correct obvious typos, rather than... well, whatever that was. –  The Chaz 2.0 Sep 28 '11 at 0:12
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1 Answer 1

How many ways are there of choosing the first ball? It could be any one of the 36 original balls, so the answer is 36.

Once that ball is out of the mix, how many ways of choosing the second ball? Well, there are no longer 36 balls in the supply; there is one fewer, so the answer is 35.

Multiply these two numbers together to find the number of different ways of choosing two different balls from the supply, with the order being significant.

Carry on multiplying (not forgetting to remove the ball you chose from the number of remaining balls) until you have selected the 6 balls.

You will have quite a big number. This is the number of different ways of choosing a sequence of 6 balls in order from a stock of 36, without repetitions.

Given that in your question the order seems to be important, this is your answer. However, for many lotteries, you wouldn't have finished yet.

For most lotteries, the order the balls were selected in is not important. You could have selected the balls in any order and the result would have been effectively the same. So you have counted all these combinations too often: 1, 2, 3, 4, 5, 6 is effectively the same as 1, 2, 3, 4, 6, 5 for example.

How many times have you counted each combination of balls? (A combination is a set of objects where the sequence the objects appear in is not important)

Well, any of the 6 balls could have been in first place, and any of the 5 remaining could have been in second.

$6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

So for each sequence without repetition, there are $720$ ways of writing it down in a different order. So divide the number of sequences by $720$ and you'll get the number of combinations of 6 balls from a stock of 36.

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