Permutations And Combinations

Question

# There are 8 students appearing for an examination, of which 3 appear in mathematics paper, and 5 in other different subjects. In how many ways can they be seated if(a) all the students appearing for mathematics paper are together.(b) the students appearing for mathematics paper are not all together.(c) no two students appearing in mathematics paper are seated together ?

Solution

Step I:  Tie the students appearing in mathematics paper together.
Number of permutations =
Mix the tie bundle with remaining 5 students to make 5 + 1 = 6.
Number of permutations =
Hence, by fundamental principle of counting, the number of ways in which the students appearing in
mathematics paper sit together = 6 x 720 = 4320.
ALTERNATIVELY:    n = 8,  r = 3.
Number of permutations = r! (n- r + 1)! = 3! (8 - 3 + 1)! = 3! 6! = 6 x 720 = 4320.
(b) Total number of ways in which 8 boys may be seated without any condition

The number of ways in which the mathematics students are sitting together = 4320.
[By part (a)]
The number of ways in which the mathematics students are not all sitting together
= 40320 - 4320 = 36000
ALTERNATIVELY:
Number of permutations = n! - [r! (n - r + 1)!] = 8! - (3! 6!) = 36000
(C)   When no two students appearing in mathematics paper are to sit together.
Arrange 5 remaining students by leaving one seat between every two students appearing in
examination other than mathematics.
Number of permutations  =

Number of places for students appearing in mathematics paper = 6.
Number of students appearing in mathematics paper = 3.
Number of permutations =
Hence, the number of arrangements when no two students appearing in mathematics paper sit                together. = 120 x 120 = 14400