A) 2400 | B) 6400 |

C) 3600 | D) 7200 |

Explanation:

Let the beds be numbered 1 to 7.

**Case 1** : Suppose Anju is allotted bed number 1.

Then, Parvin cannot be allotted bed number 2.

So Parvin can be allotted a bed in 5 ways.

After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.

So, in this case the beds can be allotted in 5´5!ways = 600 ways.

**Case 2** : Anju is allotted bed number 7.

Then, Parvin cannot be allotted bed number 6

As in Case 1, the beds can be allotted in 600 ways.

**Case 3** : Anju is allotted one of the beds numbered 2,3,4,5 or 6.

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.

Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.

Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways.

The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways

A) 1112 | B) 2304 |

C) 1224 | D) 2426 |

Explanation:

Number of questions = 5

Possibilities of choices for each question 1 to 5 respectively = 4, 4, 4, 6, 6

Reuired total number of sequences

= 4 x 4 x 4 x 6 x 6

**= 2304.**

A) 120 | B) 240 |

C) 256 | D) 360 |

Explanation:

Required number of 5 digit numbers can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits are

**5 x 4 x 3 x 2 x 1 = 120.**

A) 10000 | B) 9900 |

C) 8900 | D) 7900 |

Explanation:

Here in 100P2, P says that permutations and is defined as in how many ways 2 objects can be selected from 100 and can be arranged.

That can be done as,

$100{\mathrm{P}}_{2}$ = 100!/(100 - 2)!

= 100 x 99 x 98!/98!

= 100 x 99

**= 9900.**

A) 720 | B) 1440 |

C) 1800 | D) 3600 |

Explanation:

Given word is THERAPY.

Number of letters in the given word = 7

Number of vowels in the given word = 2 = A & E

Required number of different ways, the letters of the word THERAPY arranged such that vowels always come together is

**6! x 2! = 720 x 2 = 1440.**

A) 112420 | B) 85120 |

C) 40320 | D) 1209600 |

Explanation:

Given word is **TRANSFORMER.**

Number of letters in the given word = 11 (3 R's)

Required, number of ways the letters of the word 'TRANSFORMER' can be arranged such that 'N' and 'S' always come together is

**10! x 2!/3! **

= 3628800 x 2/6

**= 1209600**

A) 1,51,200 ways. | B) 5,04,020 ways |

C) 72,000 ways | D) None of the above |

Explanation:

In circumcstances word there are 3C's, 2S's, I, U,R, T, A, N, E

Total = 13 letters

But last letter must be N

Hence, available places = 12

In that odd places = 1, 3, 5, 7, 9, 11

Owvels = 4

This can be done in **6P4 ways **

Remaining 7 letters can be arranged in** 7!/3! x 2! ways**

Hence, total number of ways = **6P4 x 7!/3! x 2! = 360 x 5040/12 = 1,51,200 ways.**

A) 720 | B) 5040 |

C) 360 | D) 180 |

Explanation:

The number of letters in the given word RITUAL = 6

Then,

Required number of different ways can the letters of the word 'RITUAL' be arranged = 6!

**=> 6 x 5 x 4 x 3 x 2 x 1 = 720**

A) 64 | B) 63 |

C) 62 | D) 60 |

Explanation:

Given digits are 0, 4, 2, 6

Required 4 digit number should be greater than 6000.

So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.

This can be done by

1x4x4x4 = 64

Greater than 6000 means 6000 should not be there.

Hence, 64 - 1 = 63.