A) 196 | B) 186 |

C) 190 | D) 200 |

Explanation:

When at least 2 women are included.

The committee may consist of 3 women, 2 men : It can be done in $4{C}_{3}*6{C}_{2}$ ways

or, 4 women, 1 man : It can be done in $4{C}_{4}*6{C}_{1}$ways

or, 2 women, 3 men : It can be done in $4{C}_{2}*6{C}_{3}$ ways.

Total number of ways of forming the committees

= $4{C}_{2}*6{C}_{3}+4{C}_{3}*6{C}_{2}+4{C}_{4}*6{C}_{1}$

= 6 x 20 + 4 x 15 + 1x 6

= 120 + 60 + 6 =186

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.