Permutation and Combination Test Questions

Question 11

Question:  

Answer:   

Question 12

Question:   In how many ways can 6 men and 5 women be seated in a line so that no two women sit together?

Answer:    Let the women be denoted by W and the men be denoted by M.


Now the women can sit at the places marked by '.' .
We have seven places for 5 women. The women can sit in ⁷P₅ ways. Also 6 men can be arranged in 6! ways.

Question 13

Question:   In how many ways can 10 examination papers be arranged so that best and the worst are never together?

Answer:    10 papers can be arranged in 10! ways. When the best and the worst are together, the number of arrangements will be 9! x 2.
\When the best and the worst are not together, the number of arrangements will be 10! - (9!)2 = 9!(10-2) = (8) x (9!).

Question 14

Question:   In how many ways can three prizes be distributed among 4 boys when
i) No one gets more than one prize.
ii) A boy can get any number of prizes.

Answer:    i) The first prize can be given in 4 ways as one cannot get more than one prize, the remaining two prizes can be given in 3 and 2 ways respectively.
\The total number of ways = 4 x 3 x 2 = 24.

ii) As there is no restriction, each prize can be given in 4 ways.
\The total number of ways = 4³ = 64.

Question 15

Question:   How many three digit numbers are there, with distinct digits, with each digits odd?

Answer:    Here, we have to form three digit numbers with distinct digits, each digit odd.

The digits used are 1, 3, 5, 7, 9. Number of odd digits is 5. From 5 odd digits, we have to take 3 at a time. This can be done in P(5,3) ways.

\Number of three digit numbers = P(5,3) = 5 x 4 x 3 = 60

Aliter



As we require the numbers with distinct digits, repetition of digits is not allowed.

Hundred's place can be filled in 5 ways.
Ten's place can be filled in 4 ways.
Unit's place can be filled in 3 ways.

By fundamental principle of counting, required number of 3-digit numbers = 5 x 4 x 3 = 60.

Question 16

Question:   Given 6 flags of different colours, how many different signals can be generated, if a signal requires the use of two flags one below the other?

Answer:   
Number of signals obtained by 6 flags taking 2 flags at a time is P(6,2).
P(6,2) = 6 x 5 = 30

Question 17

Question:   Find n; r if

Answer:    i)






ii)





iii)









iv)







v)

Question 18

Question:   In how many ways can 6 men and 5 women sit in a row, so that women occupy the even places?

Answer:    Total number of places = 6+5 = 11

These five even places can be filled by 5 women in 5! ways and the remaining six places by 6 men in 6! ways.

Question 19

Question:   In how many ways can six children stand in a queue?

Answer:    6 children can stand in a queue in ⁶P₆ ways.
⁶P₆ = 6! = 720 ways.

Question 20

Question:   How many words with or without meanings can be formed using all the letters of the word EQUATION using each letter exactly once?

Answer:    The number of distinct letters in the word EQUATION are 8.
We have to use all the 8 letters without repetition.
Hence, the number of words formed = ⁸P₈ = P(8,8) = 8! = 40320.