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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In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
Answer: D
There are ten children. "At least one boy" is easiest through the complement.
\[ \text{total} = {}^{10}C_{4} = 210 \]
\[ \text{no boy} = {}^{4}C_{4} = 1 \]
\[ 210 - 1 = 209 \]
With only four girls there is exactly one all-girl selection, so the complement is a single case.
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The number of new words that can be formed by rearranging the letters of the word 'ALIVE' is -.
Answer: C
ALIVE has five distinct letters.
\[ 5! = 120 \text{ arrangements in all} \]
One of those is the original word itself, which is not new:
\[ 120 - 1 = 119 \]
The 120 on offer is the total count and 24 is \( 4! \) - both a step away from what is asked.
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How many four digit even numbers can be formed using the digits {2, 3, 5, 1, 7, 9}
Answer: D
A number is even only if its last digit is even, and 2 is the only even digit available.
\[ \text{last digit} = 1 \text{ choice} \]
The other three positions take the remaining five digits without repetition:
\[ 5 \times 4 \times 3 = 60 \]
\[ 1 \times 60 = 60 \]
Fixing the constrained position first is the standard tactic for these.
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A delegation of 5 members has to be formed from 3 ladies and 5 gentlemen. In how many ways the delegation can be formed, if 2 particular ladies are always included in the delegation?
Answer: A
The two particular ladies are already in, so only three more places remain and they come from the other six people (1 lady and 5 gentlemen).
\[ {}^{6}C_{3} = 20 \]
Fixing members reduces both the size of the pool and the number still to be chosen.
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