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Chill out With Permutation

Expert replies
by dtweah » Tue May 12, 2009 11:27 am
In a permutation (ordering) of the sequence 1,2,3,4,5,6 an integer n is fixed if it is in the nth position; for example 3 and 5 are fixed in the permutation 2,1,3,6,5,4. There are how many permutations of 1,2,3,4,5,6 which have exactly 3 fixed integers?
(a) 24
(b) 30
(c) 40
(d) 60
(e) 80
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Source: — Problem Solving |

by sureshbala » Tue May 12, 2009 6:48 pm
Once we fix the position of three digits in the correct position, the remaining three digits must be placed in the correct position and this can be done in only 2 ways.

For example, if we fix the digits 1,2 and 3 the arrangements possible are

1,2,3, 6.4,5

1,2,3, 6,5,4


i.e. once 1,2 and 3 are fixed 4 must be placed either in the 5th or the 6th position and for each of these two ways there will be only 1 way to fix the other 2 numbers.

Since we can fix the position of any 3 digits from 6 digits in 6C3 = 20 ways, total number of arrangements = 20 x 2 =40
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by dumb.doofus » Tue May 12, 2009 8:06 pm
sureshbala wrote:Once we fix the position of three digits in the correct position, the remaining three digits must be placed in the correct position and this can be done in only 2 ways.

For example, if we fix the digits 1,2 and 3 the arrangements possible are

1,2,3, 6.4,5

1,2,3, 6,5,4


i.e. once 1,2 and 3 are fixed 4 must be placed either in the 5th or the 6th position and for each of these two ways there will be only 1 way to fix the other 2 numbers.

Since we can fix the position of any 3 digits from 6 digits in 6C3 = 20 ways, total number of arrangements = 20 x 2 =40
Although your solution is correct.. the example that you gave is I think a typing mistake..
1) 1,2,3,6,4,5
2) 1,2,3,5,6,4 (This one your typed incorrectly)
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by sureshbala » Tue May 12, 2009 8:32 pm
dumb.doofus wrote:
sureshbala wrote:Once we fix the position of three digits in the correct position, the remaining three digits must be placed in the correct position and this can be done in only 2 ways.

For example, if we fix the digits 1,2 and 3 the arrangements possible are

1,2,3, 6.4,5

1,2,3, 6,5,4


i.e. once 1,2 and 3 are fixed 4 must be placed either in the 5th or the 6th position and for each of these two ways there will be only 1 way to fix the other 2 numbers.

Since we can fix the position of any 3 digits from 6 digits in 6C3 = 20 ways, total number of arrangements = 20 x 2 =40
Although your solution is correct.. the example that you gave is I think a typing mistake..
1) 1,2,3,6,4,5
2) 1,2,3,5,6,4 (This one your typed incorrectly)
Yup...that's a typo...thanks for the correction
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