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**there are 8 balls in a box, 3 are similar and the rest are different, how many ways can 3 balls be chosen if the order of picking out the balls is not important**

its worth 4 marks, and i do not know where to start even.

- Thread starter aerosmith
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- #1

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its worth 4 marks, and i do not know where to start even.

- #2

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Say only ball 1 appears in spot 1 (ball 2 and 3 do not appear), you have (8-3)*(8-4) = 20 possibilities of the sort. Now say ball 1 appear in spot 2, you still have 20 possibitities. Now it's easy to see why the total number of possibilties is 3*20 = 60. Now consider ball 2 and 3. You have for each 60 possibities. However since ball 1, 2 and 3 are similar, all the possibilities that occur for ball 2 and 3 have equivalents in the possibilities for ball 1, thus we only take 60 into account.

Next, use the same procedure for 2 balls appearing:In how many ways can 2 balls be rearanged with 3 spots? The awnser is 3*2 = 6. For the remaning spot, how many possibilities can you have? The awnser is 8-3 = 5. Now with using combinations, the number of possibilities is 6*5 = 30. Now since we are considering 2 balls, we are dealing with reoccuring possibilities (i.e. ball 1 inversed with ball 2 etc.). The question is, if there's 2 balls, in how many ways can they be interchanged? The awnser is simply 2*1 = 2. Since we do not wish to have reoccurences, we devided 30 by 2, giving 15.

We then consider the 3 similar balls showing up. It should be obvious that this account for only 1 possibility.

Now we have 60 + 15 + 1 = 76 as the final awnser.

- #3

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thanks for the reply, but i thought the order did not matter? why would it be 20*3 and not just 20 since order does not matter.

- #4

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- #5

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thats what i thought too......but...... if you look closer at the question, it says the order the ball was picked does not matter, that does not mean the order it will be after all 3 were picked did not matter, am i right to say that?

- #6

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Ways of choosing 3 balls from eight when three are the same : (8,3)/3!

8!/5!(3!)^2 =336/36

Probably not right, just posting my thoughts.

- #7

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How can you have non-integer number of possibilities? :tongue2:acm said:

Ways of choosing 3 balls from eight when three are the same : (8,3)/3!

8!/5!(3!)^2 =336/36

Probably not right, just posting my thoughts.

- #8

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I think 26 is the correct anwnser... If they specify that the order dosen't count, it means that different arrangements for the same balls are neglected.aerosmith said:thats what i thought too......but...... if you look closer at the question, it says the order the ball was picked does not matter, that does not mean the order it will be after all 3 were picked did not matter, am i right to say that?

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