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?
Answer: B
The two selections are independent, so their counts multiply.
\[ {}^{5}C_{2} \times {}^{6}C_{3} = 10 \times 20 = 200 \]
Order does not matter within either group, so combinations are used rather than permutations.
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Number of circular permutations of different things taken all at a time is n!.
Answer: B
The claim is false. Circular permutations of \( n \) distinct objects number
\[ (n-1)! \]
not \( n! \). In a circle there is no fixed starting point, so the \( n \) rotations of any arrangement all look the same - fixing one object's position removes that redundancy and leaves \( (n-1)! \).
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If Suraj does not want the vowels together, then in how many ways can he arrange the letters of the word 'MARKER'?
Answer: C
MARKER has six letters with R twice, and its vowels are A and E.
\[ \text{total arrangements} = \frac{6!}{2!} = 360 \]
Treating the two vowels as one block gives five units:
\[ \frac{5!}{2!} \times 2! = 60 \times 2 = 120 \text{ with the vowels together} \]
\[ 360 - 120 = 240 \]
The division by \( 2! \) throughout accounts for the identical R's.
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A student has to opt for 2 subjects out of 5 subjects for a course. Namely commerce, economics, statistics, mathematics 1 and Mathematics 2, Mathematics 2 can be offered only if mathematics 1 has also opted. The number of different combinations of two subjects which can be opted is
Answer: C
Count all pairs, then remove those that break the rule.
\[ \text{total} = {}^{5}C_{2} = 10 \]
Invalid pairs are those containing Mathematics 2 without Mathematics 1 - that is, Mathematics 2 paired with commerce, economics or statistics:
\[ 3 \text{ invalid pairs} \]
\[ 10 - 3 = 7 \]
The pair {Mathematics 1, Mathematics 2} is allowed, since the prerequisite is satisfied within the pair itself.
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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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In a class there are 20 boys and 25 girls. In how many ways can a boy and a girl be selected?
Answer: B
The two choices are independent.
\[ 20 \times 25 = 500 \]
Every one of the 20 boys can be paired with every one of the 25 girls.
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In a room there are 2 green chairs, 3 yellow chairs and 4 blue chairs. In how many ways can Raj choose 3 chairs so that at least one yellow chair is included?
Answer: C
There are \( 2 + 3 + 4 = 9 \) chairs. "At least one yellow" is easiest through the complement.
\[ \text{total} = {}^{9}C_{3} = 84 \]
\[ \text{no yellow} = {}^{6}C_{3} = 20 \]
\[ 84 - 20 = 64 \]
Counting exactly one, two and three yellows separately gives \( 3 \times 15 + 3 \times 6 + 1 = 64 \) as well - the same answer for three times the work.
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In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
Answer: C
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
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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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In how many different ways can six players be arranged in a line such that two of them, Ajeet and Mukherjee are never together?
Answer: D
As there are six players, So total ways in which they can be arranged = 6!ways =720.
A number of ways in which Ajeet and Mukherjee are together = 5!x2 = 240.
Therefore, Number of ways when they don’t remain together = 720 -240 =480.
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