Permutation under Restriction
Trending Questions
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class will plant tree, a section of class will plant trees and so on till class . There are three sections of each class. How many trees will be planted by the students?
- 11!2!2!2!−9!2!2!
- 9!2!2!2!
- 9!2!2!
- 11!2!2!2!
- the repetition is allowed is 104
- the repetition is not allowed is 9× 9P3
- the number is even and the repetition is allowed is 4500
- the number is divisible by 5 and repetition is not allowed is 952
A five letter word is to be formed such that the letters appearing in the odd positions are taken from the unrepeated letters of the word MATHEMATICS whereas the letters which occupy even places are taken from amongst the repeated letters.
- If repetition is allowed?
- If repetition is not allowed?
- 6250 , 120
- 3125 , 240
- 3125 , 120
- 1050 , 120
xA2A3A4A5A6Af211111,
where A is a positive integer has a variance of 160, then the value of A is
1. the number is between 3300 and 7200
2. the number is even
3. the number has no repeated digit.
- 48
- 144
- 288
- 576
- without any restriction is 5040
- that begins with I is 720
- that begins with I and ends with B is 120
- that neither begins with I nor ends with B is 3720
- 20
- 12
- 18
- 24
Eleven animals of a circus have to be placed in eleven cages (one in each cage). If 4 of the cages are too small for 6 big animals, then the number of the ways of caging all the animals is :
- 11P6×4!
- 11P6×5!
- 7P6×4!
- 7P6×5!
- number of cases in which the two factors are coprime is 16
- number of cases in which their H.C.F. is 3, is 32
- number of cases in which their H.C.F. 5, is 16
- number of cases in which their H.C.F is odd is 180
- 36
- 48
- 72
- 144
- 504
- 696
- 120
- 480
- 7!
- 3⋅4P3
- 4⋅3P3
- 3!⋅4!
- 20
- 12
- 18
- 24
- 720
- 3420
- 4200
- 4320
- 48
- 60
- 72
- 36
6 women and 5 men are to be seated in a row so that no 2 men can sit together. Number of ways they can be seated is
5! ×7P5
6! ×6P5
5! ×7P5
6! ×7P5
- 15
- 14
- 13
- 12
- 41×93
- 37×93
- 7×94
- 41×94
- 6
- 7
- 8
- 9
The number of 6 digit numbers that can be formed using digits 0, 1, 2, 5, 7 and 9 which are divisible by 11 and no digit is repeated are
- 360
- 1048
- 1254
- 1956
- without any restrictions is 9!2!3!
- If word starts with consonant is 4×8!2!3!
- when the word starts and end with any vowel is 7!
- If word starts with T and ends with G is 420
Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let m be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue.
Then the value of mn is
- 210
- 420
- 105
- 315