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Directions : Read the following questions carefully and choose the right answer.
1
In how many ways can the letters of the word ‘PARAGLIDING’ be arranged such that all the vowels occur together?
» Explain it
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
 
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
Five people out of whom only two can drive are to be seated in a five seater car with two seats in front and three in the rear. The people who know driving don’t sit together. Only someone who knows driving can sit on the driver’s seat. Find the number of ways the five people can be seated. 
 
» Explain it
D
Number of people who can drive = 2
 
Number of ways of selecting driver = 2C1
 
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 = 2C1 × 3
 
Number of ways of seating remaining = 3!
 
Total number of all five can be seated = 2C1 × 3 × 3!  = 36
 
Hence, correct answer is 36 

Hence, option D is correct.
3
A boy is playing a Snake & Ladder game; he is on 91 and has to get to 100 to complete the game. There is a snake on 93 and 96.  In how many ways he can complete the game, if he doesn’t want to roll the dice more than three times.
» Explain it
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.
4
8 members are to be selected from a group of 9 males and 7 females. In how many ways will the members with at most 3 females and at least 4 males be selected?
» Explain it
C
Case I: 5 males and 3 females can be selected
 
Number of ways of selection = 9C5 × 7C3 = 126 × 35 = 4410
 
Case II: 6 males and 2 females can be selected
 
Number of ways of selection = 9C6 × 7C2 = 84 × 21 = 1764
 
Case III: 7 males and 1 female can be selected
 
Number of ways of selection = 9C7 × 7C1 = 36 × 7 = 252
 
Case IV: 8 males can be selected
 
Number of ways of selection = 9C8 = 9
 
So, total number of ways of selecting the members = 4410 + 1764 + 252 + 9 = 6435 ways
 
Hence, option C is correct.
5
A chess board has rows and columns marked A to H and 1-8. Aman has a knight and a rook which he has to place on the board such that the two pieces are not in same row or column, what is total number of ways he can place the two pieces?
» Explain it
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 = 64C1 
 
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 = 49C1
 
Total number of possible ways = 49C1 × 64C1 = 3136

Hence, option B is correct.

Permutation and Combination Questions with Explanations PDF

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.

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