A) 1152 | B) 1278 |

C) 1296 | D) None of these |

Explanation:

Case I : MW MW MW MW

Case II: WM WM WM WM

Let us arrange 4 men in 4! ways, then we arrange 4 women in ways at 4 places either left of the men or right of the men. Hence required number of arrangements

A) 1000 | B) 625 |

C) 525 | D) 125 |

Explanation:

Methods for selecting 4 questions out of 5 in the first section = 5x4x3x2x1/4x3x2x1 = 5

similarly for other 2 sections also i.e 5 and 5

so total methods = 5 x 5 x 5 = 125.

A) 51/1221 | B) 42/221 |

C) 1/221 | D) 52/1245 |

Explanation:

Total Combination of getting a card from 52 cards = 52C1

Because there is no replacement, so number of cards after getting first card= 51

Now, Combination of getting an another card= 51C1

Total combination of getting 2 cards from 52 cards without replacement= (52C1)x(51C1)

There are total 4 Ace in stack. Combination of getting 1 Ace is = 4C1

Because there is no replacement, So number of cards after getting first Ace = 3

Combination of getting an another Ace = 3C1

Total Combination of getting 2 Ace without replacement=(4C1)x(3C1)

Now,Probability of getting 2 cards which are Ace = (4C1)x(3C1)/(52C1)x(51C1) = 1/221.

A) 142 | B) 175 |

C) 212 | D) 253 |

Explanation:

The first person shakes hands with 22 different people, the second person also shakes hands with 22 different people, but one of those handshakes was counted in the 22 for the first person, so the second person actually shakes hands with 21 new people. The third person, 20 people, and so on...

So,

22 + 21 + 20 + 19 + 18 + 17 + 16 + 15 + 14 + 13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1

= n(n+1)/2 = 22 x 23 /2 = 11 x 23 = 253.

A) 105 | B) 7! x 6! |

C) 7!/5! | D) 420 |

Explanation:

Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room,

^{7}C

_{1 }x

^{6}C

_{2 }x

^{4}C

_{4 }

_{ = 7 x x 1 }

_{ = 7 x 15 = 105.}

A) 215 | B) 315 |

C) 415 | D) 115 |

Explanation:

Hence, the given problem can be considered as selecting pairs of lines from the given 2 sets of parallel lines.

Therefore, the total number of parallelograms formed =

^{7}C

_{2}x

^{6}C

_{2}= 315