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Permutations and Combinations

Short Answer Type

81. How many words can be formed by using the letters of the word ‘ORIENTAL’ so that A and E always occupy the odd places?

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82.

In how many ways can 5 girls be seated in a row so that two girls Ridhi and Sanya are:

(a) always together (b) never together

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83. How many different numbers of 6-digits can be formed by using digits 4,5,6,7,8,9 (a) no digit being repeated (b) digits may be repeated?

Also, find in case (a) as to how many of them arc odd, when repetition is not allowed?

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84. How many different numbers of 5-digit can be formed by using the digits 2,3, 4, 5 and 7, without repetition, so that the number is not divisible by 5?

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85. In how many ways can three prizes be given to 10 boys when a boy may receive any number of prizes?

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86. Find the number of 4-digit numbers in which any digit may be repeated any number of times.

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87. Find the number of 4-digit numbers in which no digit is repeated.

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88. Find the number of 4-digit numbers in which at least one digit is repeated.

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Long Answer Type

89. There are three prizes to be distributed among 6 students. In how many ways can this be done when

(a) no boy gets more than one prize.

(b) there is no restriction as to the number of prizes that a boy may get.

(c) no boy sets all prizes.

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90.
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 ?

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 =