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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In Daya’s bag there are 3 books of History, 4 books of Science and 2 books of Maths. In how many ways can Daya arrange the books so that all the books of same subject are together?
Answer: D
Treat each subject as a single block, then arrange within each block.
\[ \text{blocks among themselves} = 3! = 6 \]
\[ \text{History} = 3! = 6, \quad \text{Science} = 4! = 24, \quad \text{Maths} = 2! = 2 \]
\[ 6 \times 6 \times 24 \times 2 = 1728 \]
Forgetting the \( 3! \) for the blocks themselves would give 288; the subjects can appear in any order on the shelf.
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In how many ways can three consonants and two vowels be selected from the letters of the word "TRIANGLE"?
Answer: D
TRIANGLE has consonants T, R, N, G, L and vowels I, A, E.
\[ {}^{5}C_{3} \times {}^{3}C_{2} = 10 \times 3 = 30 \]
Only the selection is asked for, not the arrangement, so combinations apply; arranging the five chosen letters would multiply this by \( 5! \).
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In Cricket League, in first round every team plays a match with every other team. 9 teams participated in the Cricket league. How many matches were played in the first round?
Answer: A
Every pair of teams meets once, and the order within a pair is irrelevant.
\[ {}^{9}C_{2} = \frac{9 \times 8}{2} = 36 \]
Counting ordered pairs would give 72, treating "A versus B" and "B versus A" as different matches.
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