Permutations And Combinations

Question
CBSEENMA11014122

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 = straight P presuperscript 3 subscript 3 space equals space 3 factorial space equals space 6
Mix the tie bundle with remaining 5 students to make 5 + 1 = 6.
Number of permutations = straight P presuperscript 6 subscript 6 equals space space 6 factorial space equals space 720.
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 
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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  = space space straight P presuperscript 5 subscript 5 space equals space 5 factorial space equals space 120.
             space space space circled times space space space space space straight R space space space space space circled times space space space space space space straight R space space space space space space circled times space space space space space space straight R space space space space space circled times space space space space space space straight R space space space space space circled times space space space space space space straight R space space space space space circled times space
           Number of places for students appearing in mathematics paper = 6.  
           Number of students appearing in mathematics paper = 3.
           Number of permutations = straight P presuperscript 6 subscript 3 space equals space fraction numerator 6 factorial over denominator 3 factorial end fraction space equals space fraction numerator 6 cross times 5 cross times 4 cross times 3 factorial over denominator 3 factorial end fraction equals 120.
           Hence, the number of arrangements when no two students appearing in mathematics paper sit                together. = 120 x 120 = 14400

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