Dear Reader,
Below are four problems on random arrangements using permutation.
Points to note:
1. Number of Permutations: Number of all permutations of n things, taken r at a time, is given by:
nPr = n(n - 1)(n - 2) ... (n - r + 1) = n!/(n - r)!
2. If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,
such that (p1 + p2 + ... pr) = n.
Then, number of permutations of these n objects is = n!/(p1!)(p2)!.....(pr!)
Question 1
In how many different ways can the letters of the word 'SERVING' be arranged?
a) 5040 b) 720 c) 120 d) 40320
Answer : a) 5040
Solution :
There are 7 different letters in the word 'SERVING'.
Therefore, the number of arrangements of the seven letters of the word = Number of all permutations of 7 letters, taken 7 at a time =
nPn = n(n - 1)(n - 2) ... (n - n + 1) = n!
Here, n = 7 then required number of ways = 7! = 5040.
Question 2
In how many different ways can any 4 letters of the word 'WORKING' be arranged?
a) 5040 b) 840 c) 24 d) 120
Answer : b) 840
Solution :
There are 7 different letters in the word 'WORKING'.
Therefore, the number of arrangements of any 4 out of seven letters of the word = Number of all permutations of 7 letters, taken 4 at a time =
nPr = n(n - 1)(n - 2) ... (n - r + 1)
Here, n = 7 and r = 4, then we have
7p4 = 7 x 6 x 5 x 4 = 840.
Hence, the required number of ways is 840.
Question 3
In how many different ways can the letters of the word 'ARRANGEMENT' be arranged?
a) 2494800 b) 4989600 c) 831600 d) none of these.
Answer : a) 2494800
Solution :
There are 11 different letters in the word 'ARRANGEMENT', in which A occurs 2 times, R occurs 2 times, n occurs 2 times and E occurs 2 times.
Therefore, the number of different arrangements of 11 letters of the word = n!/(p1!)(p2)!.....(pr!)
Here, n = 11, p1 = 2, p2 = 2, p3 = 2 and p4 = 2.
Then, 11!/2!2!2!2! = 11!/16 = 2494800.
Hence, the required number of ways is 2494800.
Question 4
In how many different ways can the letters of the word 'DIGEST' be arranged such that the vowels may appear in the even places?
a) 30 b) 720 c) 144 d) 24
Answer : c) 144
Solution :
There are 4 consonants and 2 vowels in the word DIGEST.
Out of 6 places, 3 places odd and 3 places are even.
2 vowels can arranged in 3 even places in 3p2 ways = 3 x 2 = 6 ways.
And then 4 consonants can be arranged in the remaining 4 places in 4p4 ways = 4! = 24 ways.
Hence, the required number of ways = 6 x 24 = 144.
Below are four problems on random arrangements using permutation.
Points to note:
1. Number of Permutations: Number of all permutations of n things, taken r at a time, is given by:
nPr = n(n - 1)(n - 2) ... (n - r + 1) = n!/(n - r)!
2. If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,
such that (p1 + p2 + ... pr) = n.
Then, number of permutations of these n objects is = n!/(p1!)(p2)!.....(pr!)
Question 1
In how many different ways can the letters of the word 'SERVING' be arranged?
a) 5040 b) 720 c) 120 d) 40320
Answer : a) 5040
Solution :
There are 7 different letters in the word 'SERVING'.
Therefore, the number of arrangements of the seven letters of the word = Number of all permutations of 7 letters, taken 7 at a time =
nPn = n(n - 1)(n - 2) ... (n - n + 1) = n!
Here, n = 7 then required number of ways = 7! = 5040.
Question 2
In how many different ways can any 4 letters of the word 'WORKING' be arranged?
a) 5040 b) 840 c) 24 d) 120
Answer : b) 840
Solution :
There are 7 different letters in the word 'WORKING'.
Therefore, the number of arrangements of any 4 out of seven letters of the word = Number of all permutations of 7 letters, taken 4 at a time =
nPr = n(n - 1)(n - 2) ... (n - r + 1)
Here, n = 7 and r = 4, then we have
7p4 = 7 x 6 x 5 x 4 = 840.
Hence, the required number of ways is 840.
Question 3
In how many different ways can the letters of the word 'ARRANGEMENT' be arranged?
a) 2494800 b) 4989600 c) 831600 d) none of these.
Answer : a) 2494800
Solution :
There are 11 different letters in the word 'ARRANGEMENT', in which A occurs 2 times, R occurs 2 times, n occurs 2 times and E occurs 2 times.
Therefore, the number of different arrangements of 11 letters of the word = n!/(p1!)(p2)!.....(pr!)
Here, n = 11, p1 = 2, p2 = 2, p3 = 2 and p4 = 2.
Then, 11!/2!2!2!2! = 11!/16 = 2494800.
Hence, the required number of ways is 2494800.
Question 4
In how many different ways can the letters of the word 'DIGEST' be arranged such that the vowels may appear in the even places?
a) 30 b) 720 c) 144 d) 24
Answer : c) 144
Solution :
There are 4 consonants and 2 vowels in the word DIGEST.
Out of 6 places, 3 places odd and 3 places are even.
2 vowels can arranged in 3 even places in 3p2 ways = 3 x 2 = 6 ways.
And then 4 consonants can be arranged in the remaining 4 places in 4p4 ways = 4! = 24 ways.
Hence, the required number of ways = 6 x 24 = 144.
