Aptitude Permutation And CombinationPage 2
9.  How many words can be formed by using all the letters of the word DAUGHTER so that the vowels always come together ?  (a)3420  (b)3540  (c)4320  (d)5200 

Answer is: CGiven words contains 8 different letters. When the vowels AUE are always together, we may suppose them to form an entity, treated as one letter.
then, the letters to be arranged are DGHTR(AUE)
These 6 letters can be arranged in ^{6}P_{6} = 6! = 720 ways
The vowels in the group (AUE) may be arranged in 3! = 6 ways
required number of words = (720 x 6) = 4320.
10.  In how many ways can a cricket eleven be chosen out of a batch of 15 players ?  (a)1265  (b)1276  (c)1365  (d)1375 

Answer is: CRequired number of ways = ^{15}C_{11} = ^{15}C_{(15  11)} = ^{15}C_{4}
Required number of ways = (15 x 14 x 13 x 12)/(4 x 3 x 2 x 1) = 1365
11.  In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?  (a)166  (b)5040  (c)11760  (d)86400 

Answer is: CRequired number of ways = (^{8}C_{5} x ^{10}C_{6})
Required number of ways = (^{8}C_{3} x ^{10}C_{4})
Required number of ways = [(8 x 7 x 6)/(3 x 2 x 1)] x [(10 x 9 x 8 x 7)/(4 x 3 x 2 x 1)]
Required number of ways = 11760
12.  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 included in the draw?  (a)32  (b)48  (c)64  (d)96 

Answer is: CWe may have(1 black and 2 nonblack) or (2 black and 1 nonblack) or (3 black).
Required number of ways = (^{3}C_{1} x ^{6}C_{2}) + (^{3}C_{2} x ^{6}C_{1}) + (^{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
13.  In how many different ways can the letters of the word DETAIL be arranged in such a way that the vowels occupy only the odd positions?  (a)48  (b)36  (c)60  (d)120 

Answer is: BThere are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = ^{3}P_{3} = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = ^{3}P_{3} = 3! = 6.
Total number of ways = (6 x 6) = 36.
14.  In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?  (a)63  (b)90  (c)126  (d)135 

Answer is: ARequired number of ways = ^{7}C_{5} x ^{3}C_{2} = ^{7}C_{2} x ^{3}C_{1} = [(7 x 6)/(2 x 1) x 3] = 63.
15.  How many 4letter words with or without meaning, can be formed out of the letters of the word, LOGARITHMS, if repetition of letters is not allowed?  (a)40  (b)400  (c)5040  (d)2520 

Answer is: CLOGARITHMS contains 10 different letters.
Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
Required number of words = ^{10}P_{4}
Required number of words = (10 x 9 x 8 x 7)
Required number of words = 5040.
16.  In how many different ways can the letters of the word MATHEMATICS be arranged so that the vowels always come together?  (a)10080  (b)498960  (c)120960  (d)145680 

Answer is: CIn the word MATHEMATICS, we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = 8!/(2!)(2!) = 10080.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = 4!/2! = 12.
Required number of words = (10080 x 12) = 120960.
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