Permutations and Combinations - Online Test

Q1. In a multiple choice question, there are 4 alternatives, of which one or more are correct. The number of ways in which a candidate can attempt this question is
Answer : Option B
Explaination / Solution:

Since given there are four alternatives in which one or more are correct,we have to consider the following four cases

The candidate choose 1 correct answer, 2 correct answers,3 correct answers or 4 correct answers.

1 correct answer can be chosen in 4C1 ways = 4 ways

2 correct answer can be chosen in 4C2 ways= 6 ways

3 correct answers can be chosen in  4C3ways  = 4 ways

4 correct answers can be chosen in4C4 ways = 1 way

Hence the totalnumber of ways = 4 + 6 + 4+ 1=15ways

Q2. In how many ways can the letters of the word ‘MATHEMATICS ‘be permuted so that consonants always occur together?
Answer : Option C
Explaination / Solution:

In the word 'MATHAEMATICS'  there are   7 consonants which are M-2 ,,H-1,C-1,S-1, T-2 .Since they have to occur together we treat them as a single unit.

Now this single unit together with the remaining 4 vowels which are A-2,E-1,and I-1  will account for 5 letters.

Now in these 5 letters we have A is repeating twice  these can be aarranged in   different ways.

Corresponding to each of these arrangements the consonents can be arranged in  different ways.

Hence the number of ways it can be arranged is  =75600

Q3. The number of triangles that can be formed with 6 points on a circle is
Answer : Option D
Explaination / Solution:

To form a triangle 3 non collinear points are needed.

Number of ways of selecting 3  points out of 6 can be  done in 6C3 = 20 ways.

Q4. Number of ways in which 10 different things can be divided into two groups containing 6 and 4 things respectively is
Answer : Option A
Explaination / Solution:

If there are 10 things and we have  to make them in to two groups containing 6 things and 4 things respectively , you have to select 6 to form first group , then automatically another group would have formed of 4 remaining things.

Now 6 things can be selected from 10 things in  different ways

Also we have

Q5. The number of selections of n different things taken r at a time which exclude m particular things is
Answer : Option A
Explaination / Solution:

No: of combinations of n different things taken r at at a time is given by nCr

No: of combinations of n different things taken r at a time and excluding m particular things is n - mC= C(n-m,r)

Q6. The number of arrangements of n different things taken r at a time which include a particular thing is
Answer : Option D
Explaination / Solution:

The number of arrangements of n different things taken r at a time which include a particular thing is r n-1Pr-1 = rP(n-1,r-1)

Q7. The number of arrangements of n different things taken r at a time which exclude a particular thing is
Answer : Option C
Explaination / Solution:

The number of arrangements of n different things taken r at a time which exclude a particular thing is n-1P = P(n-1,r)

Q8. 5 persons board a lift on the ground floor of an 8 storey building. In how many ways can they leave the lift?
Answer : Option A
Explaination / Solution:

Since they are boarding from ground floor and we are considering the number of ways they leave the lift ,we can consider there are 7 floor sas we exclude the ground floor)

As each of the 5 persons can leave the lift in 7 ways, required number of ways= 75

Q9.

If 12C4 + 12C=nC5  ,then n is equal to

Answer : Option B
Explaination / Solution:

We have  nCr - 1  + nCn+1Cr

Hence 12C4 + 12C13C5 ................(i)

But given 12C4 + 12C=nC5 ................(ii)

Comparing (i) and (ii) we get  n = 13

Q10. The number of ways in which a necklace can be formed by using 5 identical red beads and 6 identical white beads is:
Answer : Option A
Explaination / Solution:

Total number of beads =11

We have 6 beads are alike and next 5 beads are also alike , also since it is a necklace it can be observed from both the sides .

Therefore required number of ways=