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 |
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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 6P6 = 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 |
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Answer is: CRequired number of ways = 15C11 = 15C(15 - 11) = 15C4
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 |
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Answer is: CRequired number of ways = (8C5 x 10C6)
Required number of ways = (8C3 x 10C4)
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 |
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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) + (3C2 x 6C1) + (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
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 |
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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 = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 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 |
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Answer is: ARequired number of ways = 7C5 x 3C2 = 7C2 x 3C1 = [(7 x 6)/(2 x 1) x 3] = 63.
15. | | How many 4-letter 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 |
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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 = 10P4
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 |
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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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