Chapter 16: Permutations – Exercise 16.5
Permutations – Exercise – 16.5 – Q.1(i)
There are 12 letters in the word 'INDEPENDENCE' out of which 2 are D'S, 3 are N'S, 4 are E'S and the rest are all distinct.
so, the total number of words
.png "Permutations – Exercise – 16.5 – Q.1(i)")
= 11 × 10 × 9 × 8 × 7 × 6 × 5 = 1663200.
Permutations – Exercise – 16.5 – Q.1(ii)
There are 12 letters in the word 'INTERMEDIATE' out of which 2 are I'S, 2 are T'S, 3 are E'S and the rest are all distinct.
so, the total number of words
.png "Permutations – Exercise – 16.5 – Q.1(ii)")
= 11 × 10 × 9 × 8 × 6 × 5 × 4 × 3
= 19958400
Permutations – Exercise – 16.5 – Q.1(iii)
There are 7 letters in the word 'ARRANGE' out of which 2 are A'S, 2 are R'S, and the rest are all distinct.
So, the total number of words
.png "Permutations – Exercise – 16.5 – Q.1(iii)")
= 7 × 6 × 5 × 2 × 3
= 1260
Permutations – Exercise – 16.5 – Q.1(iv)
There are 5 letters in the word 'INDIA' out of which 2 are I'S, and the rest are all distinct.
so, the total number of
.png "Permutations – Exercise – 16.5 – Q.1(iv)")
= 60
Permutations – Exercise – 16.5 – Q.1(v)
There are 8 letters in the word 'PAKISTAN' out of which 2 are A'S, and the rest are all distinct.
So, the total number of words
.png "Permutations – Exercise – 16.5 – Q.1(v)")
= 8 × 7 × 6 × 5 × 4 × 3
= 20160
Permutations – Exercise – 16.5 – Q.1(vi)
There are 6 letters in the word 'RUSSIA' out of which 2 are S's, and the rest are all distinct.
So, the total number of words
.png "Permutations – Exercise – 16.5 – Q.1(vi)")
= 6 × 5 × 4 × 3
= 360
Permutations – Exercise – 16.5 – Q.1(vii)
There are 6 letters in the word 'SERIES' out of which 2 are S's, 2 are E's and the rest are all distinct.
so, the total number of words
.png "Permutations – Exercise – 16.5 – Q.1(vii)")
= 6 × 5 × 2 × 3
= 180
Permutations – Exercise – 16.5 – Q.1(viii)
There are 9 letters in the word 'EXERCISES' out of which 3 are E's, 2 are S's and the rest are all distinct.
So, the total number of words
.png "Permutations – Exercise – 16.5 – Q.1(viii)")
= 9 × 8 × 7 × 6 × 5 × 2
= 30240
Permutations – Exercise – 16.5 – Q.1(ix)
There are 14 letters in the word 'CONSTANTINOPLE' out of which 2 are O's, 3 are N's, 2 are T's and the rest are all distinct.
So, the total number of
.png "Permutations – Exercise – 16.5 – Q.1(ix)")
Permutations – Exercise – 16.5 – Q.2
There are 4 consonants in the word 'ALGEBRA'.
The number of ways to arrange these consonants = 4!
There are 3 vowels in the given word of which 2 are A's
The vowels can be arranged among themselves in 3!/2! ways.
Hence, the required number of arrangements
= 72
Permutations – Exercise – 16.5 – Q.3
In the word 'UNIVERSITY' there are 10 letters of which 2 are I's.
There are 4 vowels in the given word of which 2 are I's.
These vowels can be put together in 4!/2! ways.
Considering these 4 vowels as one letter there are 7 letters which can be arranged in 7! ways.
Hence, by fundamental principle of multiplication, the required number of arrangements is
= 60480.
Permutations – Exercise – 16.5 – Q.4
There are 3a's, 2b's and 4c's.
So, the number of arrangements 9!
= 9 × 4 × 7 × 5
= 1260.
Hence, the total number of arrangements are 1260.
Permutations – Exercise – 16.5 – Q.5
There are 8 letters in the word 'PARALLEL' out of which A's and 3 are L's and the rest are all distinct.
So, total number of words
= 8 × 7 × 6 × 5 × 2
= 3360
Considering all L's together and treating them as one letter we have 6 letters out of which A repeats 2 times and others are distinct. These 6 letters can be arranged in 6!/2! ways.
So, the number of words in which all L's come together
.png "Permutations – Exercise – 16.5 – Q.5(i)")
= 6 × 5 × 4 × 3 = 360
Hence, the number of words in which all L's do not come together
= 3360 - 360
= 3000.
Permutations – Exercise – 16.5 – Q.6
There are 6 letters in the word 'MUMBAI' out of which 2 are M's and the rest are all distinct.
Considering both M's together and treating as one letter we have 5 letters. These 5 letters can be arranged in 5! ways.
Hence, the total number of arrangement = 5!
= 5 × 4 × 3 × 2 × 1
= 120
Permutations – Exercise – 16.5 – Q.7
Total number of digits are = 7
There are 4 odd digits 1, 1, 3, 3 and 4 odd places (1, 3, 5, 7)
So, odd digits can be arranged in odd places in 4!/2!2! ways
The remaining 3 even digits 2, 2, 4 can be arranged in 3 even places in 3!/2! ways.
Hence, the total number of Numbers
