Permutations under restrictions
- Number of permutations of n distinct objects when a particular object is not taken in any arrangement is n-1Pr
- Number of permutations of n distinct objects when a particular object is always included in any arrangement is r.n-1Pr-1.
Without using factorials prove that nPr = n-1Pr + r.n-1Pr-1
nPr indicates permutations of n distinct objects
taken r at a time. Alternatively, we have two ways of arranging these :
One, in which a particular object is always excluded - there are n-1Pr ways of doing so.
Two, in which the particular object is always included - there are r.n-1Pr-1 ways of doing so.
Hence nPr = n-1Pr + r.n-1Pr-1
In how many ways can 4 books on Mathematics and 3 books on English be placed on a shelf so that books on the same subject always remain together?
Consider the 4 books on Mathematics as one big book and 3 books on
English as another big book. These two can be arranged in 2! ways. In each of
these arrangements, 4 books on Mathematics can be arranged among themselves in
4! ways and 3 books on English can be arranged among themselves in 3! ways.
Hence, the required number of ways = 4! 3! 2! = 288.
How many different (eight letter) words can be formed out of the letters of the word DAUGHTER so that
- the word starts with D and ends with R
- position of letter H remains unchanged
- relative position of vowels and consonants remains unaltered
- no two vowels are together?
The given word consists of 8 different letters out of which 3 are vowels and 5 are consonants.
- If the words have to start with D and end with R, then we can arrange remaining 6 letters at 6 places in 6P6 = 6! = 720 ways.
- If position of H remains unchanged, the remaining 7 letters can be arranged in 7 places in 7P7 = 7! = 5040 ways.
- The relative position of vowels and consonants remains unaltered means that vowel can take the place of vowel and consonant can take place of consonant. Now the 3 vowels can be arranged among themselves in 3! = 6 ways and the 5 consonants can be arranged among themselves in 5! = 120 ways. Thus the total number of words that can be formed = 6.120 = 720
- First let us arrange the consonants in a row. This can be done in 5P5
= 5! = 120 ways
C x C x C x C x C x
Now no two vowels are together if they are put at places marked . The 3 vowels can fill up these 6 places in 6P3 = 6.5.4 = 120 ways. Hence, the total number of words = 120.120 = 14400
- In how many ways can 7 books be arranged on a shelf if
(i) any arrangement is possible
(ii) 3 particular books must always stand together
(iii) two particular books must occupy the ends?
- Four different mathematics books, six different physics books, and two
different chemistry books are to be arranged on a shelf. How many different
arrangements are possible if
(i) the books in each particular subject must all stand together
(ii) only the mathematics books must stand together?
- In how many ways can 5 children be arranged in a row such that two boys
Ajay and Sachin
(i) always sit together
(ii) never sit together?
- When a group photograph is taken, all the nine teachers should be seated in the first row and all the eighteen students should be in second row. If the two corners are reserved for the two tallest students (interchangeable only between them), and if the middle seat of the first row is reserved for the principal, how many arrangements are possible?
- Find the number of ways in which 5 boys and 5 girls may be seated in a row so that no two girls are together.
- If ten students appear in an examination and 4 of them are appearing for mathematics and rest for 6 different subjects, in how many ways can they be seated in a row so that no copying is possible?
- Find the number of ways in which the candidates A1, A2,
A3, ..., A10 can be ranked if
(i) A1 and A2 are next to each other
(ii) A1 is always above A2.
- How many words can be formed out of the letters of the word ARTICLE so that vowels occupy even places?
- How many words can be formed out of the letters of the word ORIENTAL so that A and E occupy odd places?
- How many words beginning and ending with a consonant can be formed by using the letters of the word EQUATION?
- How many different five letter words starting with a vowel can be formed from the letters of the word EQUATION?
- How many arrangements can be formed by the letters of the word VOWELS if
(i) there is no restriction
(ii) each word begins with S
(iii) each word begins with S and ends with E
(iv) all vowels come together
(v) all consonants come together?
- The letters of the word TRIANGLE are arranged in such a way that vowels and consonants remain together. How many different arrangements will be obtained?
- How many 6 digit telephone numbers can be constructed with the digits 0, 1, 2, ..., 9 if each number starts with 35 and no digit appears more than once?
- How many four digit numbers divisible by 4 can be made with the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?
Answers1. (i) 5040 (ii) 720 (iii) 240 2. (i) 207360 (ii) 8709120
3. (i) 48 (ii) 72 4. 2. 6! 8!
5. 86400 6. 604800
7. (i) 2! 9! = 725760 (ii) 10!/2 = 1814400
8. 144 9. 8640
10. 4320 11. 4200
12. (i) 720 (ii) 120 (iii) 24 (iv) 240 (v) 144
13. 1440 14. 1680 15. 24