A) 109 | B) 190 |

C) 901 | D) 910 |

Explanation:

Number of positive integral solutions =

=

= 190

A) 7920 | B) 74841 |

C) 14874 | D) 10213 |

Explanation:

ST candidates vacancies can be filled by 5**C**3 ways = 10

Remaining vacancies are 5 that are to be filled by 12

=> 12**C**5 = (12x11x10x9x8)**/**(5x4x3x2x1) = 792

Total number of filling the vacancies = 10 x 792 = 7920

A) F is third to the left of B | B) F is second to the right of B |

C) B is an immediate neighbour of D | D) All of the above |

Explanation:

From the given information, the circular arrangement is

Here F is second to the right of B and the remaning all are wrong.

A) 43929 | B) 59049 |

C) 15120 | D) 0 |

Explanation:

Number of words with 5 letters from given 9 alphabets formed =

Number of words with 5 letters from given 9 alphabets formed such that no letter is repeated is =

Number of words can be formed which have at least one letter repeated = -

= 59049 - 15120

= 43929

A) 5040 | B) 3650 |

C) 4150 | D) 2520 |

Explanation:

Total number of letters in the word ABYSMAL are 7

Number of ways these 7 letters can be arranged are 7! ways

But the letter is repeated and this can be arranged in 2! ways

=> Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.

A) 186 | B) 144 |

C) 136 | D) 120 |

Explanation:

In given word OLIVER there are 3 vowels E, I & O. These can be arranged in only one way as dictionary order E, I & O.

There are 6 letters in thegiven word.

First arrange 3 vowels.

This can be done in 6C3 ways and that too in only one way.(dictionary order E, I & O)

Remaining 3 letters can be placed in 3 places = 3! ways

Total number of possible ways of arranging letters of OLIVER = 3! x 6C3 ways

= 6x5x4 = 120 ways.