Chapter 16: Permutations – Exercise 16.4
Permutations – Exercise – 16.4 – Q.1
There are 4 vowels and 3 consonants in the word 'FAILURE'
We have to arrange 7 letters in a row such that consonants occupy odd places. There are 4 odd places (1, 3, 5, 7). There consonants can be arranged in these 4 odd places in 4P3 ways.
Remaining 3 even places (2, 4, 6) are to be occupied by the 4 vowels. This can be done in 4P3 ways.
Hence, the total number of words in which consonants occupy odd places = 4P3 × 4P3
= 4 × x 3 × 2 × 1 × 4 × 3 × 2 × 1
= 24 × 24
= 576.
Permutations – Exercise – 16.4 – Q.2
There are 7 letters in the word 'STRANGE', including 2 vowels (A, E) and 5 consonants (S, T, R, N, G).
(i) Considering 2 vowels as one letter, we have 6 letters which can be arranged in 6p6 = 6! ways A,E can be put together in 2! ways.
Hence, required number of words
= 6i × 2!
= 6 × 5 × 4 × 3 × 2 × 1 × 2
= 720 × 2
= 1440.
(ii) The total number of words formed by using all the letters of the words 'STRANGE' Is 7p7 = 7!
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040.
So, the total number of words in which vowels are never together
= Total number of words – Number of words in which vowels are always together
= 5040 - 1440
= 3600
(iii) There are 7 letters in the word 'STRANGE'. out of these letters 'A' and 'E' are the vowels. There are 4 odd places in the word 'STRANGE'. The two vowels can be arranged in 4p2 ways. The remaining 5 consonants can be arranged among themselves in 5p5 ways.
The total number of arrangements
Permutations – Exercise – 16.4 – Q.3
There are 6 letters in the word 'SUNDAY'. The total number of words formed with these 6 letters is the number of arrangements of 6 items, taken all at a time, which is equal
to 6p6 = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.
If we fix up D in the beginning, then the remaining 5 letters can be arranged in 5P5 = 5! ways.
so, the total number of words which begin with D = 5!
= 5 × 4 × 3 × 2 × 1 = 120.
Permutations – Exercise – 16.4 – Q.4
There are 4 vowels and 4 consonants in the word 'ORIENTAL'. We have to arrange 8 leeters in a row such that vowels occupy odd places. There are 4 odd places (1, 3, 5, 7). Four vowels can be arranged in these 4 odd places in 4! ways. Remaining 4 even places (2, 4, 6, 8) are to be occupied by the 4 consonants.
This can be done in 4! ways.
Hence, the total number of words in which vowels occupy odd places = 4! × 4!
= 4 × 3 × 2 × 1 × 4 × 3 × 2 × 1 = 576.
Permutations – Exercise – 16.4 – Q.5
There are 6 letters in the word 'SUNDAY'. The total number of words formed with these 6 letters is the number of arrangements of 6 items, taken all at a time, which is equal to 6p6 = 6!
= 6 × 5 × 4 × 3 × 2 × 1 = 720.
If we fix up N in the begining, then the remaining 5 letters can be arranged in 5p5 = 5! ways so, the total number of words which begin which N = 5!
= 5 × 4 × 3 × 2 × 1
= 120
if we fix up N in the begining and Y at the end, then the remaining 4 letters can be arranged in 4p4 = 4! ways.
So, the total number of words which begin with N and end with Y = 4! = 4 × 3 × 2 × 1 = 24.
Permutations – Exercise – 16.4 – Q.6
There are 10 letters in the word 'GANESHPUR1'. The total number of words formed is equal to 10p10 = 10!
(i) If we fix up G in the begining, then the remaining 9 letters can be arranged in 9p9 = 9! ways
(ii) If we fix up P in the begining and I at the end, begining 8 letters can be arranged in 8p8 = 8!.
(iii) There are 4 vowels and 6 consonants in the word 'GANESHPURI'.
Considering 4 vowels as one letter,
We have 7 letters which can be arranged in 7p7 = 7! ways.
A,E,U,I can be put together in 4! ways.
Hence, required number of words = 7! × 4!.
(iv) We have to arrange 10 letters in a row such that vowels occupy even places. There are 5 even places (2, 4, 6, 8, 10), 4 vowels can be arranged in these 5 even places in 5p4 ways.
Remaining 5 odd places (1, 3, 5, 7, 9) are to be occupied by the 6 consonants.
This can be done in 6C5 ways.
Hence, the total number of words in which vowels occupy even places = 5p4 × 6P5
Permutations – Exercise – 16.4 – Q.7
(i) There are 6 letters in the word 'VOWELS'. The total number of words formed with these 6 letters is the number of arrangements of 6 items, taken all at a time, which is equal to
6p6 = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
(ii) If we fix up E in the begining then the remaining 5 letters can be arranged in
5p5 = 5! = 5 × 4 × 3 × 2 × 1 = 120 ways
(iii) If we fix up 0 in the begining and L at the end, the remaining 4 letters can be arranged in
4p4 = 4! = 4 × 3 × 2 × 1 = 24.
(iv) There are 2 vowels and 4 consonants in the word 'VOWELS'. Considering 2 vowels as one letter, we have letters which can be arranged in 5p5 = 5! ways.
O, E can be put together in 2! ways.
Hence, required number of
words = 5! × 2!
= 5 × 4 × 3 × 2 × 1 × 2 × 1
= 120 × 2
= 240
(v) There are 2 vowels and 4 consonants in the word 'VOWELS'.
Considering 4 consonants as one letter, we have 3 letters which can be arranged in 3p3 = 3! ways. U, W, L, S can be put together in 4! ways.
Hence, required number of words in which all consonants come together = 3! × 4!
= 3 × 2 × 4 × 3 × 2
= 144.
Permutations – Exercise – 16.4 – Q.8
We have to arrange 7 letters in a row such that vowels occupy even places.
There are 3 even places (2, 4, 6). Three vowels can be arranged in these 3 even places in 3! ways.
Remaining 4 odd places (1, 3, 5, 7) are to be occupied by the 4 consonants. This can be done in 4! ways.
Hence, the total number of words in which vowels occupy even places = 3! × 4!
= 3 × 2 × 4 × 3 × 2 = 144
Permutations – Exercise – 16.4 – Q.9
Let two husbands A, B be selected out of seven males in = 7C2 ways. excluding
their wives, we have to select two ladies C,D out of remaining 5 wives is = 5C2 ways.
Thus, number of ways of selecting the players for mixed double is = 7C2 × 5C2
= 21 × 10
= 210
Now, suppose A chooses C as partner (B will automatically go to D) or A chooses 0 as partner (B will automatically go to C) Thus we have, 4 other ways for teams.
Required number of ways = 210 × 4 = 840
Permutations – Exercise – 16.4 – Q.10
m men can be seated in a row in mpm = m! ways.
Now, in the (m + 1) gaps n women can be arranged in m+1pn ways.
Hence, the number of ways in which no two women sit together
Hence, proved
Permutations – Exercise – 16.4 – Q.11
(i) MONDAY has 6 letters with no repetitions, so
Number of words using 4 letters at a time with no repetitions = 6p4
= 6!/2!
= 360
(ii) Number of words using all 6 letters at a time with no repetitions = 6p6
= 6 × 5 × 4 × 3 × 2 × 1
= 720
(iii) Number of words using all 6 letters, starting with vowels
= 2,5p5
= 2 × 5 × 4 × 3 × 2 × 1
= 240
Permutations – Exercise – 16.4 – Q.12
There are 8 letters in the word 'ORIENTAL'. The total number of three letter words is the number of arrangements of 8 items, taken 3 at a time, which is equal to
= 336.
Hence, the total number three letter words are 336.