Formula for P(n,r)
The number of permutations of n different things taken r at a time is given by
P(n, r) = n. (n -1). (n -2) ... (n -r +1) = n! / (n -r)!
Note. n Pn = n (n -1) ... (n - n +1) = n (n -1) ... 1 = n!
Illustrative Examples
Example
Find the values of (i)5P5
(ii) 5P3 (iii)60P48
(iv)60P2.
Solution
- 5P5 = 5. 4. 3. 2. 1 = 120
- 5P3 = 5. 4. 3 = 60
- 60P48 = 60! / (60 -48)! = 60! / 12!
- 60P2 = 60. 59 = 3540
Example
Find n if
- 5P4 :nP5 = 1 : 2
- nP4 :n -1P3 = 9 : 1
- P (n, 4) = 2 P (5, 3)
Solution
- nP4 :nP5 = 1 : 2
=> [n (n -1)(n -2)(n -3)] / [n (n -1)(n -2)(n -3)(n -4)] = 1/2
- nP4 :n -1P3 = 9 : 1
=> [n (n -1)(n -2)(n -3)]/[(n -1)(n -2)(n -3)] = 9/1
=> n = 9
- P (n, 4) = 2 P (5, 3) = 2. 5. 4. 3 = 5. 4. 3 . 2
=> n (n -1)(n -2)(n -3) = 5 . 4. 3. 2
=> n = 5
Example
- In how many ways can the letters of the word WONDERFUL be arranged?
- Ten students participate in a debate. In how many ways can the first
three prizes be won?
- Find the number of signals which can be given with 5 flags of different
colours hoisted one above the other, by using any number of flags.
Solution
- Since there are 9 distinct letters in WONDERFUL, the required number of
permutations is
9P9 = 9! = 9. 8. 7. 6. 5. 4. 3. 2. 1 = 362880
- The given problem is similar to one of filling 3 places with any 3 of 10
distinct objects. This can be done in 10P3 ways. Hence required
number is
10P3 = 10. 9. 8 = 720
- When 1 flag is used, number of signals = 5P1 = 5
When 2 flags are used, number of signals = 5P2 = 5. 4 = 20
When 3 flags are used, number of signals = 5P3 = 5. 4. 3 = 60
When 4 flags are used, number of signals = 5P4 = 5. 4. 3. 2 = 120
When 5 flags are used, number of signals = 5P5 = 5. 4. 3. 2. 1 = 120
Hence total number of signals = 5 +20 +60 +120 +120 = 325
Example
- Find the number of permutations of n different things taken r at a time
when each thing may be repeated any number of times in any permutation.
- In how many ways can three prizes be given to 20 boys when a boy may
receive any number of prizes?
- Find the sum of all three digit numbers formed by using odd digits 1, 3,
5, 7, 9 with repetition of digits allowed.
Solution
- The first place may be filled with any of n things. After the first place
has been filled up, the second place can also be filled in n ways since we
are not prevented from repeating the same thing. When the first two places
have been filled in n x n ways, the third place can also be filled up in n ways and so on.
By fundamental principle of association, the r places can be filled in
(n x n x ... r times) ways, i.e. nr ways.
Hence the required number of permutations = nr
- The first prize can be given in 20 ways, and then second prize can also
be given in 20 ways, since repetition is allowed. Similarly third prize can
also be given in 20 ways. Thus by principle of association, required number
of ways = 20 x 20 x 20 = 8000
- Since any of 5 digits can be placed at any place, there are 5 x 5 x 5 = 125 such numbers.
Now consider the digits in units place. 25 numbers end with digit 1; 25
numbers end with digit 3, and so on. Hence sum of all units digits of 125 numbers
= 25(1 +3 +5 +7 +9) = 625.
Similarly sum of tens digits of 125 numbers is 625, and sum of hundreds
digits of 125 numbers is 625.
Hence sum total of these 125 numbers is 625 (1 +10 +100) = 625 x 111 = 69375
Example
Find the number of divisors of the number 36000.
Solution
Factorising the given number, we find 36000 = 25
. 3². 5³. This means that any divisor of 36000 is of the type 2a. 3b.
5c where a can take values 0, 1, 2, 3, 4, 5; b can take values 0,
1, 2; c can take values 0, 1,2, 3. Hence number of divisors is 6 x 3 x 4 = 72.
Note that both 1 and 36000 are counted among 72 divisors.
Exercise
- Find the values of
(i) P (4, 3)
(ii) P (65, 15)
(iii)7P5
(iv)10P3
- Prove that
(i) P (10, 3) = P (9, 3) +3. P (9, 2)
(ii) P(n, n) = 2. P (n, n -2)
(iii) P (n, n) = P (n, n -1)
(iv) P(n, r) = (n -r +1). P (n, r -1)
(v) 1 +1. P1 +2. P2 +... + n. Pn = Pn =
1 where Pm = mPm
- Find n if
(i) 5 P (4, n) = 6 P (5, n -1)
(ii)n+1P3 = nP4
(iii) P (n -1, 3) : P (n +1, 3) = 5 : 12
(iv) P (n, 4) = 5040
- Find the value of r if
(i) P (11, r) = P (12, r -1)
(ii) P (9, r) = 3024
(iii) P (15, r -1) : P (16, r -2) = 3 : 4
- If r
s
n, prove that P (n, s) is
divisible by P (n, r).
[Hint. P(n, s) / P(n, r) = ... = a whole number]
- In how many ways can the letters of following words be permuted?
(i) CATS
(ii) DELHI
(iii) JAIPUR
(iv) HEXAGON
- In how many ways can you take 3 letters of above words and arrange into
words with no letter being repeated?
- (i) In how many ways can 5 people be seated on a sofa if there are only 3 seats available?
(ii) In how many ways can 3 people sit if there are 5 vacant seats?
- In how many ways can 5 children stand in a queue?
- In how many ways can 6 women draw water from 6 taps if no tap remains unused?
- Seven candidates are contesting an election. In how many ways can their
names be listed on the ballot paper?
- There are three different rings to be worn in four fingers with atmost
one in each finger. In how many ways can this be done? If the wedding ring is
to be worn on ring finger only, then what are the number of ways?
- In how many ways can captain and vice captain be chosen out of 11 players?
- In how many ways can 8 boys and 7 girls be photographed if
(i) girls are to sit on chairs in a row and the boys stand in a row behind them
(ii) girls sit on even numbered chairs and boys sit on odd numbered chairs
and there are 15 chairs in a row
(iii) there are 15 chairs in a row and there is no restriction?
- How many numbers consisting of five different digits can be made from the
digits 1, 2, 3, ..., 9 if
(i) the number must be odd
(ii) the first two digits of each number are even?
- Solve problem 15 if repetition of digits is allowed.
- Find the number of all five-digit numbers with distinct digits.
- How many natural numbers less than 1000 can be formed with the digits 1, 2, 3, 4 and 5 if
(i) no digit is repeated
(ii) repetition of digits is allowed?
- There are 3 candidates and 5 voters. In how many ways can the votes be
given? If 3 candidates are included in 5 voters and they vote for themselves
only, then how many ways of giving votes are possible?
- There are 8 true-false statements in a question paper. How many sequences of answers are possible?
- Find the number of ways in which 5 boys and 5 girls can be seated alternately in a row?
- Find the number of divisors of the number 8800. How many of these are even?
Answers
1. (i) 24 (ii) 65! / 50!
(iii) 2520 (iv) 720
3. (i) 3 (ii) 5
(iii) 8 (iv) 10
4. (i) 9 (ii) 4 (iii) 14
6. (i) 24 (ii) 120 (iii) 720 (iv) 5040
7. (i) 24 (ii) 60 (iii) 120 (iv) 210
8. (i) 60 (ii) 60
9. 120
10. 720
11. 5040
12. 24, 6
13. 110
14. (i) 8!.7! (ii) 8!.7! (iii) 15!
15. (i) 8400 (ii) 2520
16. (i) 32805 (ii) 11664
17. 27216
18. (i) 85 (ii) 155
19. (i) 243 (ii) 9
20. 256
21. 28800
22. 36, 30 [Hint. For even, power of 2 should be minimum 1]