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Number of permutations of ‘n’ things, taken ‘r’ at a time,

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Number of permutations of ‘n’ things, taken ‘r’ at a time, [#permalink]

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New post 18 Apr 2013, 14:28
Hi,
Can anyone please give me example for this -

Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement

= r x P(n-1 r-1)
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Re: Number of permutations of ‘n’ things, taken ‘r’ at a time, [#permalink]

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New post 19 Apr 2013, 02:46
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you have 6 numbered balls - from 1 to 6. You need to select 3 balls such that ball that is numbered '4' is always selected. Additionally, you need to arrange them in bags A B and C.

In essence you are selecting 2 balls from the remaining 5 balls and subsequently, arranging the 3 balls in bags A B and C.
Cases:
1.) you can place the ball#4 in A , and arrange the remaining 2 balls in B and C - in 5P2 ways OR
2.) you can place the ball#4 in B , and arrange the remaining 2 balls in A and C - in 5P2 ways OR
3.) you can place the ball#4 in C , and arrange the remaining 2 balls in B and A - in 5P2 ways

so the total cases are 3 x 5P2 i.e. r x (n-1)P(r-1)


In short, select 2 balls out of remaining 5 in 5C2 ways and arrange the 3 balls in 3! ways. Hence, 3! x 5C2 = 3 x 5P2 = r x(n-1)P(r-1)
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Re: Number of permutations of ‘n’ things, taken ‘r’ at a time, [#permalink]

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New post 19 Apr 2013, 07:57
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iwantto wrote:
Hi,
Can anyone please give me example for this -

Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement

= r x P(n-1 r-1)



Think of the formula in this way:

You have n total things, say 10. You need to select r out of those, say 4, such that one of those, say a red ball, is always included.

If you have to select and arrange 4 things out of 10, you use 10p4. Now since the red ball must always be included, we need to select only 3 out of 9 things.

We select and atrrange 3 things out of 9 in 9P3 ways i.e. (n-1)P(r-1). Now the problem leftover is where do you put the red ball? There are 4 places to put the red ball. Before the first thing, before the second thing, before the third thing and after the third thing. Hence, you multiply 9P3 by 4. You get

r * (n-1)P(r-1)
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Re: Number of permutations of ‘n’ things, taken ‘r’ at a time,   [#permalink] 19 Apr 2013, 07:57
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