Permutations and Combinations - Online Test

Q1. A fair dice is rolled n times. The number of all the possible outcomes is
Answer : Option C
Explaination / Solution:

each time there are 6 possibilities, therefore for n times there are 6n  possibilities.

Q2. The sum of all the numbers which can be formed by using the digits 1 , 3 , 5 , 7 ,9 all at a time and which have no digit repeated is
Answer : Option A
Explaination / Solution:

First we will fix any one digit in a fixed position .Then we have the remaining 4 digits can be arranged in 4! different ways.

Which means each of the five digits can appear in each of the five places in  4! times.

Hence the sum of the digits in each position is 

Now to find the sum  of these numbers formed we have to consider the place values  for these digits

So the sum of all the numbers which can be formed by using the digits 1 , 3 , 5 , 7 ,9=

 


Q3. 4 boys and 4 girls are to be seated in a row. The number of ways in which this can be done, if the boys and girls sit alternately, is
Answer : Option B
Explaination / Solution:

there are 4 boys and 4 girls and the row can start either with a boy or girl, therefore the number of ways are 4! x 4! x 2

Q4. The letters of the word ‘SOCIETY’ are arranged in such a manner that the vowels and consonants occur alternately, the number of different words so obtained is
Answer : Option B
Explaination / Solution:

there are four consonants and 3 vowels, therefore 3! * 4!

Q5. There are 10 true-false questions. The number of ways in which they can be answered is
Answer : Option A
Explaination / Solution:

you can either choose true or false, therefore for 10 questions you will have 210 possibilities.

Q6. The number of different ways in which a man can invite one or more of his 6 friends to dinner is?
Answer : Option B
Explaination / Solution:

He can invite any one  friend in 6C1 ways= 6 ways:

He can invite any two friends in 6C2 ways = 15 ways

He can invite any three friends in 6C3 ways =20 ways

He can invite any 4 friends in 6C4ways = 15 ways

He can invite any 5 friends in 6C5 ways = 6 ways

He can invite  all the 6 friends in 6C6 ways= 1 way.

Since any one of these could happen total possibilities are, 6+15+20+15+6+1 = 63.


Q7. A lady arranges a dinner party for 6 guests .The number of ways in which they may be selected from among 10 friends if 2 of the friends will not attend the party together is
Answer : Option A
Explaination / Solution:

Let the friends be A,B,C,D,E,F,G,H,I , J and assume A and B will not battend together

Case 1 : Both of them will not attend the party : Now we have to select the 6 guests from the remaing 8 members 

 Then no of ways is 8C6 = 28 ways.

Case 2 : Either of them are selected for the party :Now we have to select the 5 guests from the remaing 8 members and  one from A and B

Then the no of ways = 2C1 x 8C5 =112 ways.

Therefore total number of ways is 28+ 112 = 140 ways,


Q8. In how many ways can a mixed doubles tennis game be arranged from a group of 10 players consisting of 6 men and 4 women
Answer : Option C
Explaination / Solution:

A team of 4 players are to be selected.

2 out of 6 men can be done in 6C2 ways. 2 out of 4 women can be done in 4C2ways. 

So the number of ways to select 4 players is 6C24C2= 90.

Now we can arrange these people to form  mixed doubles.

 If M1,M2,W1,W2, are the 4 members selected then one team can be chosen as   in 2 different ways 

Therefore the required number of arrangements= 90x2 = 180.


Q9. The number of significant numbers which can be formed by using any number of the digits 0, 1, 2, 3, 4 but using each not more than once in each number is
Answer : Option C
Explaination / Solution:

Case 1: for 5 digit numbers Ten Tens thousands place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied in 4 ways : and the tens place in 3 ways and hundreds place in 2 ways and the thousands place in 1 way. Hence the total number of ways is 4x1x4x2x3= 96.

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Case 2: for 4 digit numbers thousands place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied in 4 ways : and the tens place in 3 ways and hundreds place in 2 ways . Hence the total number of ways is 4x4x2x3= 96.

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Case 3: for 3digit numbers hundreds place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied in 4 ways : and the tens place in 3 ways  Hence the total number of ways is 4x4x3= 48

Case 4: for 2 digit numbers Tens place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied. Hence the total number of ways is 4x4=16.

Case 5: For single digit in 4 ways .

Hence 96 + 96 + 48+ 16+4 = 260


Q10. Numbers greater than 1000 but not greater than 5000 are to be formed with the digits 0, 1, 2, 3, 5, allowing repetitions, the number of possible numbers is
Answer : Option D
Explaination / Solution:

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One's place can be occupied by any of the  5 numbers, tens place by any of the 5 numbers and hundreds place in 5 ways since repetition is allowed. But thousands place can be occupied by 2,3,1, only since the required number should be greater than 1000 and less than 5000.Hence  total number of arrangement=3x5x5x5=375