Permutation and Combination is one of important topic in Quantitative Aptitude Section. In Permutation and Combination – Quiz 3 article candidates can find questions with an answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Permutation and Combination – Quiz 3 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.
Q1
A committee has 5 men and 6 women. What are the number of ways of selecting 2 men and 3 women from the given committee?
The number of ways to select two men and three women = [latex]^{5}{C}_{2} \times ^{6}{C}_{3}[/latex]
= [latex]\frac {(5 \times 4)}{(2 \times 1)}[/latex] [latex]\times[/latex] [latex]\frac {(6 \times 5 \times 4)}{(3 \times 2)}[/latex]
= 200
Q2
What are the number of ways to select 3 men and 2 women such that one man and one woman are always selected?
The number of ways to select three men and two women such that one man and one woman are always selected = Number of ways selecting two men and one woman from men and five women
= [latex]^{4}{C}_{2} \times ^{5}{C}_{1}[/latex]
= [latex]\frac {(4 \times 3)}{(2 \times 1)}\times 5[/latex]
= 30 ways
Q3
A committee has 5 men and 6 women. What are the number of ways of selecting a group of eight persons?
Total number of persons in the committee = 5 + 6 = 11
Number of ways of selecting group of eight persons = [latex]^{11}{C}_{8} \times ^{11}{C}_{3}[/latex] = [latex]\frac {(11 \times 10 \times 9)}{(3 \times 2)}[/latex] = 165 ways
Q4
There are 18 stations between Hyderabad and Bangalore. How many second class tickets have to be printed, so that a passenger can travel from any station to any other station?
The total number of stations = 20
From 20 stations we have to choose any two stations and the direction of travel (i.e., Hyderabad to Bangalore is different from Bangalore to Hyderabad) in [latex]^{20}{P}_{2}[/latex] ways.
[latex]^{20}{P}_{2}[/latex] = 20 * 19 = 380.
Q5
A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most -
Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 [latex]\times[/latex] 6 [latex]\times[/latex] 6 = 216. Of these attempts, one of them is a successful attempt.
Maximum number of unsuccessful attempts = 216 - 1 = 215.