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?
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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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?
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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In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
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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How many three letter words are formed using the letters of the word TIME?
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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Using all the letters of the word "NOKIA", how many words can be formed, which begin with N and end with A?
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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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?
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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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?
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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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?
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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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?
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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In how many different ways can four books A, B, C and D be arranged one above another in a vertical order such that the books A and B are never in continuous position?
Answer: B
Count all orders, then remove those with A and B together.
\[ \text{total} = 4! = 24 \]
Treating A and B as one block gives three units, and the pair can be either way round:
\[ 3! \times 2! = 6 \times 2 = 12 \]
\[ 24 - 12 = 12 \]
With only four books, exactly half the arrangements keep A and B apart.
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