Permutations & Combinations
Important Instructions
How many combinations are possible while selecting four letters from the word ‘SMOKEJACK’ with the condition that ‘J’ must appear in it?
Answer: D
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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
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A selection is to be made for one post of principal and two posts of vice-principal amongst the six candidates called for the interview only two are eligible for the post of principal while they all are eligible for the post of vice-principal. The number of possible combinations of selectees is
Answer: D
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Find the number of rectangles and squares in an 8 by 8 chess board respectively.
Answer: B
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In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
Answer: B
The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
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A boy has nine trousers and 12 shirts. In how many different ways can he select a trouser and a shirt?
Answer: D
The boy can select one trouser in nine ways.
The boy can select one shirt in 12 ways.
The number of ways in which he can select one trouser and one shirt is 9 * 12 = 108 ways.
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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
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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
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Find the number of ways of arranging the letters of the words DANGER, so that no vowel occupies odd place.
Answer: C
The given word is DANGER. Number of letters is 6. Number of vowels is 2 (i.e., A, E). Number of consonants is 4 (i.e., D,N,G,R). As the vowels cannot occupy odd places, they can be arranged in even places. Two vowels can be arranged in 3 even places in 3P2 ways i.e., 6. Rest of the consonants can arrange in the remaining 4 places in 4! ways. The total number of arrangements is 6 x 4! = 144.
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Without repetition, using digits 2, 3, 4, 5, 6, 8 and 0, how many numbers can be made which lie between 500 and 1000?
Answer: D
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