Aptitude Permutation And CombinationPage 3
17. | | In how many different ways can the letters of the word OPTICAL be arranged so that the vowels always come together? | | (a)120 | | (b)720 | | (c)4320 | | (d)2160 |
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Answer is: BThe 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.
18. | | In how many different ways can the letters of word AUCTION be arranged in such a way that the vowels always come together ? | | (a)30 | | (b)48 | | (c)144 | | (d)576 |
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Answer is: DThe word AUCTION has 7 different letters,
when the vowels AUIO are always together, they cn be supposed to form one lettter
Then, we have to arrange the letters CTN(AUIO).
Now, 4 letters can be arranged in = 4! = 24 ways.
The vowels (AUIO) can be arranged among themselves in = 4! = 24 ways.
Required number of ways = 24 x 24 = 576
19. | | A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be inclined in the draw ? | | (a)32 | | (b)48 | | (c)64 | | (d)96 |
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Answer is: CWe may have (1 black and 2 non – black) or (2 black and 1 non-black) or (3 black)
Required number of ways = (3C1 x 6C2) or (3C2 x 6C1) or (3C3)
Required number of ways = [3 x (6 x 5)/(2 x 1)] + [(3 x 2)/(2 x 1) x 6] + 1
Required number of ways = 45 + 18 + 1 = 64
20. | | In how many different ways can the letters of the word MACHINE be arranged so that the vowels may occupy only the odd positions ? | | (a)210 | | (b)576 | | (c)1728 | | (d)3456 |
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Answer is: BThere are 7 letters in the given word, out of which there are 3 vowels and 4 consonants.
Let us mark the positions to be filled up as follows:
(1)(2)(3)(4)(5)(6)(7)
Now, 3 Vowels can be placed at any of the three places. Out of the four marked 1, 3, 5, 7
Number of ways of arranging the vowels = 4P3 = (4 x 3 x 2) = 24
Also, the 4 consonants at the remaining 4 positions may be arranged in = 4P4 = 4! = 24 ways.
Required number of ways = (24 x 24) = 576
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