Using the Permutation and Combination Rules

sweetkendra

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Feb 24, 2008
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Ok I need to know if I am doing this right. Your input will be greatly appreciated.

There are 7 seals and 8 dolphins entered in an aquatic obedience show. How many ways could you select a final group to judge if it must consist of 3 seals and 3 dolphins.

nPr= n!/(n-r)!
nCr= n!/(n-r)!r!

7C3 * 8C3= 7!/(7-3)!3! * 8!/(8-3)!3! = 7!/4!3! * 8!/5!3! = 7*6*5/4*3*2 * 8*7*6/5*4*3 = 210/24 * 336/60 = 49

although the answer that I got is not any of the answere on my paper. Please help.
 
sweetkendra said:
There are 7 seals and 8 dolphins entered in an aquatic obedience show. How many ways could you select a final group to judge if it must consist of 3 seals and 3 dolphins.
You have six slots to fill. Order doesn't matter but, for convenience, let's say that the first three slots will be filled with seals and the last three with dolphins. So you have six slots to fill:

. . . . .___ * ___ * ___ * ___ * ___ * ___

You have seven choices for the first slot:

. . . . ._7_ * ___ * ___ * ___ * ___ * ___

Once you have picked one seal, you have six choices remaining for the second slot, and then five choices for the third slot:

. . . . ._7_ * _6_ * _5_ * ___ * ___ * ___

By similar reasoning, you have eight choices for the fourth slot, seven for the fifth, and six for the sixth:

. . . . ._7_ * _6_ * _5_ * _8_ * _7_ * _6_

Multiplying, you get 70560 ways of choosing the animals in the order SSSDDD (where "S" stands for a seal and "D" stands for a dolphin). But this ordering is not necessary so, to get rid of the permutation-type selection I've done above, we need to divide by 3! 3! (one "three factorial" for each of the two permuted sets, to "divide out" the implied ordering, which we don't need). This will give you an answer which matches what you should have gotten with the combination-formula below.

sweetkendra said:
nPr= n!/(n-r)!
nCr= n!/(n-r)!r!

7C3 * 8C3= 7!/(7-3)!3! * 8!/(8-3)!3! = 7!/4!3! * 8!/5!3! = 7*6*5/4*3*2 * 8*7*6/5*4*3 = 210/24 * 336/60 = 49
First, let me say "Thank you!" for defining your terms and showing your work! You make a tutor's work so much easier!
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Looking at your last line above, I think perhaps some factors were dropped...?

. . . . .\(\displaystyle \left(\frac{7!}{3!\,4!}\right)\,\left( \frac{8!}{5!\,3!}\right)\, =\, \left(\frac{7\times 6\times 5}{3\times 2\times 1}\right)\,\left( \frac{8\times 7\times 6}{3\times 2\times 1}\right)\)

. . . . . . . . . . . . . . . . . .. . .\(\displaystyle =\, \left( \frac{7\times 5}{1} \right)\, \left( \frac{8\times 7}{1}\right)\)

...which does not equal 49. :shock:

Note that, with the arithmetic corrected, your method and the slot-filling method displayed above both give the same answer. :D

Eliz.
 
:) Thank you so much Eliz. What you showed me help me a great deal. I got the answer of 1960. Your help is greatly appreciated. I have to take my test in the morning, I am not looking forward to it at all.

Thanks again,

Karen
 
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