Thursday, 22 December 2016
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From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
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| 2. |
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
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| 3. |
In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
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| 4. |
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
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| 5. |
In how many ways can the letters of the word 'LEADER' be arranged?
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| 6. |
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
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| 7. |
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
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| 8. |
In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
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| 9. |
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?
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| 10. |
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?
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| 11. |
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?
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| 12. |
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?
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| 13. |
In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?
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| 14. |
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
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Answer:
1. Answer: Option D
Explanation:
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
 Required number of ways | = (7C3 x 6C2) + (7C4 x 6C1) + (7C5) | |||||||||||
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| = (525 + 210 + 21) | ||||||||||||
| = 756. |
2.Answer: Option C
Explanation:
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
* Required number of ways = (120 x 6) = 720.
3. Answer: Option D
Explanation:
In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
| Number of ways arranging these letters = | 7! | = 2520. |
| 2! |
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
| in | 5! | = 20 ways. |
| 3! |
* Required number of ways = (2520 x 20) = 50400.
4.Answer: Option C
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
| = (7C3 x 4C2) | |||||||||
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| = 210. |
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
| Number of ways of arranging 5 letters among themselves | = 5! |
| = 5 x 4 x 3 x 2 x 1 | |
| = 120. |
* Required number of ways = (210 x 120) = 25200.
5.Answer: Option C
Explanation:
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
| * Required number of ways = | 6! | = 360. |
| (1!)(2!)(1!)(1!)(1!) |
6. Answer: Option D
Explanation:
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
| * Required number of ways | = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) | |||||||||||||||||||
| = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2) | ||||||||||||||||||||
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| = (24 + 90 + 80 + 15) | ||||||||||||||||||||
| = 209. |
7. Answer: Option D
Explanation:
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
8. Answer: Option C
Explanation:
| Required number of ways | = (8C5 x 10C6) | |||||||
| = (8C3 x 10C4) | ||||||||
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| = 11760. |
9. Answer: Option C
Explanation:
We 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) | |||||||||||||
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| = (45 + 18 + 1) | ||||||||||||||
| = 64. |
10. Answer: Option C
Explanation:
There 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.
11. Answer: Option A
Explanation:
| Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = | ( | 7 x 6 | x 3 | ) | = 63. |
| 2 x 1 |
12. Answer: Option C
Explanation:
'LOGARITHMS' contains 10 different letters.
| Required number of words | = Number of arrangements of 10 letters, taking 4 at a time. |
| = 10P4 | |
| = (10 x 9 x 8 x 7) | |
| = 5040. |
13. Answer: Option C
Explanation:
In 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! | = 10080. |
| (2!)(2!) |
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
| Number of ways of arranging these letters = | 4! | = 12. |
| 2! |
* Required number of words = (10080 x 12) = 120960.
14. Answer: Option B
Explanation:
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.
