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Permutations and Combinations

Short Answer Type

71. How many 3-digit numbers can be formed by using digits 1 to 9 if no digit is repeated?

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72.

How many 4-digit numbers are there if no digit is repeated?

How many numbers are there between 1000 and 9999 so that no digit is repeated?

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73. How many words with or without meaning can be formed using all the letters of the word ‘EQUATION’ using each letter exactly once?

Additional part: How many of these words begin with E and end with N?

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74. How many words with or without meaning can be formed using all the letters of the word ‘MONDAY’ assuming, that no letter is repeated, if 4 letters are used at a time.

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75. How many words with or without meaning can be formed using all the letters of the word ‘MONDAY’ assuming, that no letter is repeated, if all letters are used at a time.

rightwards double arrow

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76. How many words with or without meaning can be formed using all the letters of the word ‘MONDAY’ assuming, that no letter is repeated, if all letters are used but the first letter is a vowel?

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77. How many words, with or without meaning, can be formed with the letters of word ‘MONDAY’ assuming, that no letter is repeated, if 4 letters are used but the first letter is a vowel ?

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78.

How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

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79.

How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that no two consonants are together.

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80. In how many ways can 6-girls and 4 boys be seated in a row so that no two boys are together?

No two boys are together.

Step I: Arrange 6 girls in a row by leaving one seat between every two girls.

Number of permutations of arranging girls = 6! = 720 ...(i)

   $circled times$   G $circled times$ G $circled times$ G $circled times$ G $circled times$ G $circled times$ G    $circled times$
1 2 3 4 5 6 7

Step II:  There are 7 places for the boys to sit in order that no two of them are together.

Number of boys = 4

   $rightwards double arrow$ n = 7, r = 4

∴ Number of permutations of arranging boys = $straight P presuperscript 7 subscript 4 space equals space fraction numerator 7 factorial over denominator 3 factorial end fraction space equals space 7 cross times 6 cross times 5 cross times 4 space equals space 840$ ...(ii)

Hence, from (i) and (ii),

The number of ways in which they may be seated so that no two boys are together

= 720 x 840 = 604800

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