## Permutation And Combination

Question 1 |

( ^{6}P_{4}) /2! | |

$4\times 3^5$ | |

( ^{40}P_{6}) /6! | |

None of the above |

The subsequent places can be filled in 3 ways each.

Hence, the number of ways =4 *3 *3 *3 *3 *3 = 4 *3^{5}

Question 2 |

5040 | |

40 | |

2520 | |

400 |

Required number of words

= Number of arrangements of 10 letters, taking 4 at a time.

= ^{10}P_{4}

= (10 x 9 x 8 x 7)

= 5040.

Question 3 |

15 | |

5 | |

10 | |

20 |

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

∴ Required number of numbers = (1 x 5 x 4) = 20.

Question 4 |

360 | |

320 | |

720 | |

540 |

If the first digit is 6, the other digits can be arranged in 6! /2! = 360 ways.

If the first digit is 7, the other digits can be arranged in 6!/(2!$\times$2!)=180 ways.

Thus required possibilities for n,

= 360 +180 = 540 ways.

Question 5 |

19 | |

18 | |

16 | |

17 |

^{n}C_{2} = 153.

On solving we get,

=> n =-17 and n =18.

It cannot be negative so,

n = 18 is the answer.

Question 6 |

None of these | |

1 | |

^{6}P_{5} | |

5 |

Question 7 |

432 | |

360 | |

384 | |

470 |

**1, 2, 3, 4, 5, 6, 7, 8**, 9, 10, 11, 12.

The various combinations of chairs that ensure that no two men are sitting together are listed.

(1, 3, 5,...), The fourth chair can be 5,6,10,11 or 12, hence 5 ways.

(1, 4, 8, ...), The fourth chair can be 6,10,11 or 12 hence 4 ways.

(1, 5, 8, ...), the fourth chair can be 10,11 or 12 hence 3 ways.

(1, 6, 8,...), the fourth chair can be 10,11 or 12 hence 3 ways.

(1,8,10,12) is also one of the combinations.

Hence, 16 such combinations exist.

In case of each these combinations we can make the four men inter arrange in 4! ways.

Hence, the required result =16$\times$4!= 384.

Question 8 |

480 | |

720 | |

360 | |

5040 |

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Therefore Required number of ways = (120 x 6) = 720.

Question 9 |

^{th}card?

2 ^{10} *3^{3} | |

4 ^{2} *3^{3} | |

4 *3 ^{4} | |

2 ^{10} |

The remainder of the number on the next card when divided by 4 can have 3 possible vales (except the one occurred earlier).

For each value on the card the remainder can have 3 possible values.

The total number of possible sequences is: 4 *3^{4}.

Question 10 |

^{12}C_{3} *^{4}C_{4} | |

^{16}C_{7} *7! | |

^{12}C_{4} *^{4}C_{3} *7! | |

^{11}C_{4} *^{4}C_{3} |

^{12}C_{4} ways.

3 vowels can be selected in ^{4}C_{3} ways.

Therefore, total number of groups each containing 4 consonants and 3 vowels,

= ^{12}C_{4} *^{4}C_{3} Each group contains 7 letters, which can be arranging in 7! ways.

Therefore required number of words,

= ^{12}C_{4} *^{4}C_{3} *7!