Permutations & Combinations Questions and Answers

Q: In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

The men and women are drawn from separate pools.\[ {}^{7}C_{5} \times {}^{3}C_{2} = 21 \times 3 = 63 \]\( {}^{7}C_{5} = {}^{7}C_{2} = 21 \), which is the quicker way to evaluate it.

Q: Three dice, each having six faces numbered 1 to 6, are rolled. What is the number of possible outcomes in which at least one die shows the number 2?

"At least one" is easiest through the complement.\[ \text{total outcomes} = 6^{3} = 216 \]\[ \text{no die shows 2} = 5^{3} = 125 \]\[ 216 - 125 = 91 \]Counting exactly one, exactly two and exactly three 2's gives \( 75 + 15 + 1 = 91 \) as well, for considerably more work.

Q: On a shelf, 2 books of Geology, 2 books of Sociology and 5 of Economics are to be arranged in such a way that the books of any subject are to be together. Find in how many ways can this be done?

Treat each subject as a block, then arrange within each.\[ \text{blocks} = 3! = 6 \]\[ \text{Geology} = 2!, \quad \text{Sociology} = 2!, \quad \text{Economics} = 5! \]\[ 6 \times 2 \times 2 \times 120 = 2880 \]The three subject blocks can themselves be ordered in any of six ways along the shelf.

Q: There are 5 tasks and 5 persons, Task 1 cannot be assigned to either person 1 or person 2, Task 2 must be assigned to either person-3 or person-4 every person is to be assigned one task, In how many ways can the assignment be done?

Deal with the most restricted assignment first.Task 2 must go to person 3 or person 4:\[ 2 \text{ ways} \]Task 1 then cannot go to person 1 or 2, and one of persons 3 and 4 is already used, so two people remain eligible for it:\[ 2 \text{ ways} \]The other three tasks go to the remaining three people freely:\[ 3! = 6 \]\[ 2 \times 2 \times 6 = 24 \]Assigning the constrained tasks first keeps the later counts independent of which choice was made.

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61.

A briefcase has a number-lock system containing a combination of 3 digits (Each digit can be of numbers 0 to 8). If the correct combination is unknown, how much maximum time would be required to open the bag if each “trial” of combination takes 3 seconds?

A.45.23 minutes
B.36.45 minutes
C.60.34 minutes
D.90.15 minutes

Answer: B

Each digit runs from 0 to 8, giving nine choices, and repetition is allowed.

\[ 9 \times 9 \times 9 = 729 \text{ combinations} \]

\[ 729 \times 3 = 2187 \text{ seconds} \]

\[ \frac{2187}{60} = 36.45 \text{ minutes} \]

The digits go up to 8, not 9, so it is \( 9^{3} \) and not \( 10^{3} \).

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62.

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

A.63
B.90
C.126
D.45

Answer: A

The men and women are drawn from separate pools.

\[ {}^{7}C_{5} \times {}^{3}C_{2} = 21 \times 3 = 63 \]

\( {}^{7}C_{5} = {}^{7}C_{2} = 21 \), which is the quicker way to evaluate it.

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63.

A group consists of 4 men, 6 women and 5 children. In how many ways can 2 men , 3 women and 1 child selected from the given group?

A.300
B.600
C.750
D.900

Answer: B

The three selections are independent, so their counts multiply.

\[ {}^{4}C_{2} \times {}^{6}C_{3} \times {}^{5}C_{1} \]

\[ = 6 \times 20 \times 5 = 600 \]

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64.

In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

A.266
B.5040
C.11760
D.86400

Answer: C

The men and the women are chosen from separate pools, so the counts multiply.

\[ {}^{8}C_{5} \times {}^{10}C_{6} \]

\[ = 56 \times 210 = 11{,}760 \]

Using \( {}^{8}C_{3} \) and \( {}^{10}C_{4} \) gives the same values, since choosing whom to include is equivalent to choosing whom to leave out.

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65.

How many three letter words are formed using the letters of the word TIME?

A.12
B.20
C.16
D.24

Answer: D

TIME has four distinct letters, and order matters in a word.

\[ {}^{4}P_{3} = 4 \times 3 \times 2 = 24 \]

The corresponding combination \( {}^{4}C_{3} = 4 \) would count only which three letters are used, not how they are arranged.

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66.

Using all the letters of the word "NOKIA", how many words can be formed, which begin with N and end with A?

A.3
B.6
C.24
D.120

Answer: B

The first and last letters are fixed as N and A, leaving O, K and I for the three middle positions.

\[ 3! = 6 \]

Without the restriction all five letters could move, giving \( 5! = 120 \).

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67.

Three dice, each having six faces numbered 1 to 6, are rolled. What is the number of possible outcomes in which at least one die shows the number 2?

A.36
B.81
C.91
D.116

Answer: C

"At least one" is easiest through the complement.

\[ \text{total outcomes} = 6^{3} = 216 \]

\[ \text{no die shows 2} = 5^{3} = 125 \]

\[ 216 - 125 = 91 \]

Counting exactly one, exactly two and exactly three 2's gives \( 75 + 15 + 1 = 91 \) as well, for considerably more work.

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68.

A locker in bank has 3 digit lock. Mahesh forgot his password and was trying all possible combinations. He took 6 seconds for each try. The problem was that each digit can be from 0 to 9. How much time will be needed to by Mahesh to try all the combinations?

A.90 minutes
B.120 minutes
C.60 minutes
D.100 minutes

Answer: D

Each of the three digits can be any of 0 to 9, so the digits are independent.

\[ 10 \times 10 \times 10 = 1000 \text{ combinations} \]

\[ 1000 \times 6 = 6000 \text{ seconds} = \frac{6000}{60} = 100 \text{ minutes} \]

Repetition is allowed here - nothing stops a code such as 777 - so it is \( 10^{3} \) rather than \( 10 \times 9 \times 8 \).

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69.

On a shelf, 2 books of Geology, 2 books of Sociology and 5 of Economics are to be arranged in such a way that the books of any subject are to be together. Find in how many ways can this be done?

A.3846
B.2880
C.900
D.90

Answer: B

Treat each subject as a block, then arrange within each.

\[ \text{blocks} = 3! = 6 \]

\[ \text{Geology} = 2!, \quad \text{Sociology} = 2!, \quad \text{Economics} = 5! \]

\[ 6 \times 2 \times 2 \times 120 = 2880 \]

The three subject blocks can themselves be ordered in any of six ways along the shelf.

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70.

There are 5 tasks and 5 persons, Task 1 cannot be assigned to either person 1 or person 2, Task 2 must be assigned to either person-3 or person-4 every person is to be assigned one task, In how many ways can the assignment be done?

A.6
B.12
C.24
D.144

Answer: C

Deal with the most restricted assignment first.

Task 2 must go to person 3 or person 4:

\[ 2 \text{ ways} \]

Task 1 then cannot go to person 1 or 2, and one of persons 3 and 4 is already used, so two people remain eligible for it:

\[ 2 \text{ ways} \]

The other three tasks go to the remaining three people freely:

\[ 3! = 6 \]

\[ 2 \times 2 \times 6 = 24 \]

Assigning the constrained tasks first keeps the later counts independent of which choice was made.

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