Permutation & combination
Q1.
The number of parallelogram that can be formed from a set 3 parallel lines intersecting another set of 4 parallel lines .
(a) 18
(b) 15
(c) 10
(d) 21
Solution.
Apply formula nC2 x mC2
3C2 x
4C2 = 3 x 6 = 18
Total number of parallelograms are 18
Answer = 18
Solution
Q2.
How many numbers can be formed by the digits 1,2,3,4,3,1,2 by placing the odd digits at odd places .
(a) 12
(b) 20
(c) 18
(d) 19
There are four odd digits and four odd places and 3 even digits with 3 even places
4!/2! x 2!
x
3!/2!
= 18
Answer = 18
Q3.
How many number between 300 and 3000 can be formed with the digits 0,1,2,3,4 and 6, no digit repeated in any time .
(a) 180
(b) 160
(c) 200
(d) 175
Solution.
(i) 3 digit number can be formed by = 3 x 5P2 ways (hundredth place can be filled up by 3 ways 3,4,6, and rest two places
can be filled up by 5P2)
= 60
(ii) 4 digit can be formed by = 2 x 5P3 ways = 120
(2 way to fill thousand place 1 or 2 and rest places can be filled by 5P3)
Total number that can be
developed are 180
Answer = 180
Solution
Q4.
Six persons A,B,C,D,E and F are to be seated at a circular table . In how many ways can this be done , if A must always have either
B or C on his right and B must always have either C or D on his right .
(a) 19 ways
(b) 12 ways
(c) 16 ways
(d) 18 ways
Solution.
As per question, we get the following possible ways .
(i)
ABC, D,E,F given (4-1)! = 3! = 6 ways
(ii) ABC, C,E,F giving (4-1)! = 3! = 6 ways
(iii) AC, BD, E,F giving (4-1)! = 3! = 6 ways .
Hence the total number of ways = 6 + 6 + 6 = 18 ways
Answer = 18
Solution
Q5.
In how many ways can 9 people be arranged at a round table , so that two particular persons be together .
(a) 3!
(b)2!
(c) 7!
(d) 9!
Solution.
7!
= 2 person sit together consider as a single unit , and in circular arrangement this will be (8-1)!
Solution
