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 

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 

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 

Answer is: CWe may have (1 black and 2 non â€“ black) or (2 black and 1 nonblack) or (3 black)
Required number of ways = (^{3}C_{1} x ^{6}C_{2}) or (^{3}C_{2} x ^{6}C_{1}) or (^{3}C_{3})
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 

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 = ^{4}P_{3} = (4 x 3 x 2) = 24
Also, the 4 consonants at the remaining 4 positions may be arranged in = ^{4}P_{4} = 4! = 24 ways.
Required number of ways = (24 x 24) = 576
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