Permutation and Combination Quiz 1

Ans: There are totally 6 letters of which N appears twice.

Total arrangements = 6!/2!

and the arrangements where N’s are together are 5!

Ans: 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 vowels as one letter, the number of letters becomes 8 which can be arranged as \frac{8!}{2!} = 20160 

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

\frac{4!}{2! \times 2!} = 6 

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

Ans: Number of people who can drive = 2

Number of ways of selecting driver = _{1}C^{2}\textrm{}

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 =  _{1}C^{2}\textrm{} \times 3

Number of ways of seating remaining = 3!

Total number of all five can be seated = _{1}C^{2}\textrm{} \times 3 \times 3!   = 36

Ans: 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 two rolls 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

Ans: Case I: 5 males and 3 females can be selected

Number of ways of selection =_{5}C^{9}\textrm{} \times _{3}C^{7}\textrm{} = 126 \times 35 = 4410 

Case II: 6 males and 2 females can be selected

Number of ways of selection = _{6}C^{9}\textrm{} \times _{2}C^{7}\textrm{} = 84 \times 21 = 1764

Case III: 7 males and 1 female can be selected

Number of ways of selection =  _{7}C^{9}\textrm{} \times _{1}C^{7}\textrm{}= 36 \times 7 = 252

Case IV: 8 males can be selected

Number of ways of selection = _{8}C^{9}\textrm{} = 9 [\latex]

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

Ans: There are a total of 15 students who are available for selection and Mukesh has to select 5 from these 15. Hence, the total number of ways in which this can be done will be

_{5}C^{15}\textrm{} = 3003

However, this includes all possible cases. We have to select the total number of cases where there is at least 1 girl in the team.

So we can do

Cases with at least 1 girl = Total cases – Cases where there is no girl in the team.

Cases when no girl is selected in the team = _{5}C^{7}\textrm{}

(The team must be selected from the boys only)

  _{5}C^{7}\textrm{} = 21

Hence, the required number of cases = 3003 – 21 = 2982

Ans: The Lays packets can be arranged among themselves in 6! ways, the cookies packet in 4! ways and chocolate packet in 2! ways and there are three groups which can be arranged in 3! ways.

∴ Reqd. number of arrangements = 6! 4! 2! 3! = 720 \times 24 \times 2 \times 6 = 207360

Ans: If three people do not want to sit next to each other, they will occupy alternate chairs, i.e. the first, third and fifth chairs. They can be arranged on these 3 chairs in 3! Ways.

The remaining two people can be arranged on the second and the fourth chairs in 2! Ways.

∴ The total number of arrangements = 3! × 2! = 12

Ans: First we select 8 people out of 10 , no of ways = _{8}C^{10}\textrm{}

Now, out of the selected 8 people we first seat one person on any of the two chairs of any one side, no of ways = 2

Now the rest of the seats can be filled in 7! Ways

Total number of ways = _{8}C^{10}\textrm{} \times 2 \times 7!= \frac{10!}{8}