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### Solved examples for permutations of n objects when r objects are taken at a time

:: Questions ::
1. Find the values of
• P(6, 6)
On comparing with P(n, r),
n = 6
r = 6
P(6, 6) = n! / (n - r)!
= 6! / (6 - 6)!
= 6! / 0!
= 720/1             .................         [since 0! = 1]

Therefore P(6, 6) = 720
• P(4, 2)
On comparing with P(n, r), we get
n = 4
r = 2
P(4, 2) = n! / (n - r)!
= 4!/ (4 - 2)!
= 4! / 2!
= 24/2
= 12

Therefore P(4, 2) = 12
• P(7, 2)
On comparing with P(n, r), we get
n = 7
r = 2
P(7, 2) = 7! / (7 - 2)!
= 7! / 5!
= 5040 / 120
= 42

Therefore the number of permutations is 42.
• P(70, 2)
On comparing with P(n, r), we get
n = 70
r = 2
Using the formula, P(n, r) = n! / (n - r)!
P(70, 2) = 70! / (70 - 2)!
= 70! / 68!
Now we know that 70! = 70 x 69 x 68!, therefore
= (70 x 69 x 68!) / 68!
= 70 x 69               ...............       [the 68! from numerator is cancelled by 68! in the denominator]
= 4830

Therefore the number of permutations is 4830

2. Find 'n' in the following equations:
• P(n, 6) = P(n, 5)
Using the formula P(n, r) = n! / (n - r)!, we get
n! / (n - 6)! = n! / (n - 5)!
By cross multiplication, we get,
n! / (n - 6)! * (n - 5)! / n! = 1
(n! * (n - 5)!) / ((n - 6)! * n!)  = 1
n! in the numerator is cancelled by n! in the denominator,
(n - 5)! / (n - 6)! = 1
We know that (n - 5)! = (n - 5)(n - 6)!, therefore
((n - 5)(n - 6)!) / (n - 6)! = 1
The (n - 6)! in the numerator is cancelled by the (n - 6)! in the denominator, therefore
(n - 5) = 1
Adding 5 to both sides, we get
n = 6
Therefore, the answer is n = 6
• P(n, 5) = P(n-1, 4)
Using the formula P(n, r) = n! / (n - r)!, we get
n! / (n - 5)! = (n - 1)! / (n - 1 - 4)!
n! / (n - 5)! = (n - 1)! / (n - 5)!
By cross multiplication, we get,
n! / (n - 5)! *  (n - 5)! / (n - 1)! = 1
(n! * (n - 5)!) / ((n - 5)! * (n - 1)!) = 1
The (n - 5)! present in the numerator and denominator of the fraction above get cancelled out by each other, so we get
n! / (n - 1)! = 1
We know that n! = n(n - 1)!, so we get
n(n - 1)! / (n - 1)! = 1
The (n - 1)! in the numerator and denominator of the fraction above get cancelled out by each other, so we get
n = 1
Therefore, the value of n is 1.
3. Fifteen students participate in a contest. In how many ways can the first three prizes be distributed?

Total number of students, n = 15
Total number of prizes to be distributed, r = 3
Total number of permutations, P(n, r) = n! / (n - r)!
Therefore, P(15, 3) = 15! / (15 - 3)!
= 15! / 12!
We know that 15! = 15 x 14 x 13 x 12!, therefore
= 15 x 14 x 13 x 12! / 12!
The 12! present in the numerator and denominator get cancelled out, so we get,
= 15 x 14 x 13
= 2730
Therefore there are 2730 different ways in which 3 prizes can be distributed among 15 students.
4. Find the number of signals which can be hosted using five flags of different color all at a time.

The number of flags present, n = 5
Number of flags hosted at a time, r = 5
Therefore total number of permutations, P(n, r) = n! / (n - r)!
Therefore,
P(5, 5) = 5! / (5 - 5)!
P(5, 5) = 5! / 0!                 ................        [since o! = 1]
P(5, 5) = 5! = 120
Therefore the total number of ways in which all the 5 different flags can be hoisted are 120.

Hope the workings are clear. For any clarification, please post the problem in the comments. Thanks.