There are 30 people in a party. If everyone is to shake hands with one another, how many hand shakes are possible?
Answer: D
A handshake involves two people and the order does not matter, so the count is a combination.
\[ {}^{30}C_{2} = \frac{30 \times 29}{2} = 435 \]
Each of the 30 people shakes 29 hands, but that counts every handshake twice, hence the division by 2.
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A box contains 4 black, 3 red and 6 green marbles. 2 marbles are drawn from the box at random. What is the probability that both the marbles are of the same color?
Answer: B
The box holds \( 4 + 3 + 6 = 13 \) marbles.
\[ \text{total ways} = {}^{13}C_{2} = 78 \]
\[ \text{same colour} = {}^{4}C_{2} + {}^{3}C_{2} + {}^{6}C_{2} = 6 + 3 + 15 = 24 \]
\[ P = \frac{24}{78} \]
The three colours are mutually exclusive, so their counts add before dividing.
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A box contains 2 white, 3 black and 5 red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?
Answer: D
There are \( 2 + 3 + 5 = 10 \) balls, 7 of them not black. Use the complement.
\[ \text{total} = {}^{10}C_{3} = 120 \]
\[ \text{no black} = {}^{7}C_{3} = 35 \]
\[ 120 - 35 = 85 \]
Counting one, two and three black balls separately gives \( 3 \times 21 + 3 \times 7 + 1 = 85 \) as well.
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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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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?
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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A person can go from place “P” to “Q” by 7 different modes of transport, but is allowed to return back to “P” by any mode other than the one used earlier. In how many different ways can he complete the entire journey?
Answer: A
Seven modes are available going out, and the return must differ from the one used.
\[ 7 \times 6 = 42 \]
The restriction removes exactly the seven journeys that would repeat a mode.
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How many words can be formed by using all letters of word ALIVE.
Answer: D
ALIVE has five letters, all different.
\[ 5! = 120 \]
No letter repeats, so there is no division to make - every arrangement is distinct.
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How many 3-letter words can be formed out of the letters of the word ‘CORPORATION’, if repetition of letters is not allowed?
Answer: B
CORPORATION has eleven letters, but repetition of letters is not allowed in the word formed, so only the distinct letters may be used:
\[ C,\ O,\ R,\ P,\ A,\ T,\ I,\ N \rightarrow 8 \text{ distinct} \]
Order matters in a word, so this is a permutation:
\[ {}^{8}P_{3} = 8 \times 7 \times 6 = 336 \]
The repeated O's and R's contribute nothing extra, since no letter may be used twice.
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In how many different ways can the letters of the word ‘GEOMETRY’ be arranged so that the vowels always come together?
Answer: C
GEOMETRY has eight letters with E appearing twice. Its vowels are E, O, E.
Treat the three vowels as one block, giving six units to arrange (the block plus G, M, T, R, Y):
\[ 6! = 720 \]
Within the block the three vowels can be ordered in
\[ \frac{3!}{2!} = 3 \text{ ways} \]
since the two E's are identical.
\[ 720 \times 3 = 2160 \]
Treating the E's as distinguishable would give \( 720 \times 6 = 4320 \) - double-counting every arrangement, because swapping the two E's changes nothing visible.
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In a group containing 6 cows and 4 buffalos, 4 livestock are to be selected in such a way that at least 1 cow should always be present. How many way of doing that are possible?
Answer: A
There are ten animals in all. "At least one cow" is easiest through the complement.
\[ \text{total} = {}^{10}C_{4} = 210 \]
\[ \text{no cow} = {}^{4}C_{4} = 1 \]
\[ 210 - 1 = 209 \]
With only four buffalos there is exactly one all-buffalo selection, which is what makes the complement so small.
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