Important for :

1

B

In the word “PARAGLIDING” there are 11 letters in which there are 4 vowels (i.e. 2 A’s and 2 I’s) and 7 consonants (i.e 2 G’s and each of P, R, L, D, N)

Considering vowel as one letter, the number of letters becomes 8 which can be arranged as

Considering vowel as one letter, the number of letters becomes 8 which can be arranged as

8! | = | 40320 | = 20160 |

2! | 2 |

Vowel A and I appear twice , so vowels can be arranged as

4! | = | 24 | = 6 |

(2! × 2!) | 4 |

Hence the required number of ways in which the letters of the word “PARAGLADING” be arranged so that all the vowels occur together = 20160 × 6 = 120960

Hence, option B is correct.

2

D

Number of people who can drive = 2

Number of ways of selecting driver = 2C_{1}

The other person who knows driving can be seated only in the rear three seats in 3 ways

Total number of ways of seating the two persons = 2C_{1} × 3

Number of ways of seating remaining = 3!

Total number of all five can be seated = 2C_{1} × 3 × 3! = 36

Hence, correct answer is 36

Hence, option D is correct.

3

C

91 --92 — 93 — 94 — 95 — 96 — 97 — 98 — 99 — 100

Total position advance needed = 100 – 91 = 9

One roll of dice can’t complete the game.

If he completes in two roll of dice.

Possible dice throws are – (3&6), (4&5), (5&4), (6&3)

But (5&4) will bring the token on 96, so this is rejected.

If he completes the game in three roll of dices

First dice reading options are 1,3,4,6

After checking all option and rejecting those in which token reaches on 93 or 96

Possible dice throws are (1,2,6), (1,3,5),(1,5,3), (1,6,2) ;

(3,1,5),(3,3,3),(3,4,2),(3,5,1);

(4,2,3),(4,3,2),(4,4,1)

(6, 1, 2), (6, 2, 1)

Total number of ways = 16

Hence, option C is correct.

Hence, option C is correct.

4

C

Case I: 5 males and 3 females can be selected

Number of ways of selection = ^{9}C_{5} × ^{7}C_{3} = 126 × 35 = 4410

Case II: 6 males and 2 females can be selected

Number of ways of selection = ^{9}C_{6} × ^{7}C_{2} = 84 × 21 = 1764

Case III: 7 males and 1 female can be selected

Number of ways of selection = ^{9}C_{7} × ^{7}C_{1} = 36 × 7 = 252

Case IV: 8 males can be selected

Number of ways of selection = ^{9}C_{8} = 9

So, total number of ways of selecting the members = 4410 + 1764 + 252 + 9 = 6435 ways

Hence, option C is correct.

5

B

As shows in the image a knight and a rook has to be placed, but not in the same row or column.

Let us select any box out of 64 for placing knight, no of ways = ^{64}C_{1}

Now, row 6 and column c can’t be used to place rook. Remaining boxes = 64 – (8 + 7) = 49

The rook can be place in any of 49 boxes, no of ways = ^{49}C_{1}

Total number of possible ways = ^{49}C_{1} × ^{64}C_{1} = 3136

Hence, option B is correct.

There are various ways in which objects from a set can be selected, generally without replacement to form subsets. The selection of subsets is known as permutation when order of selection is a factor, a combination is when the order is not a factor.

Permutation and combination is an important topic for any competitive exam. At Smartkeeda we have covered this topic in depth, so you can understand and gain mastery over this topic.

Practice daily at Smartkeeda with in-depth explanation to gain an understanding of the topic and improve your score.

Regards

**Team Smartkeeda**

Permutation and combination is an important topic for any competitive exam. At Smartkeeda we have covered this topic in depth, so you can understand and gain mastery over this topic.

Practice daily at Smartkeeda with in-depth explanation to gain an understanding of the topic and improve your score.

Regards