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1. In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all person of the same nationality sit together?
These three person can be arranged themselves in 3! Ways.
8 Indians can be arranged themselves in 8! Way.
4 American can be arranged themselves in 4! Ways.
4 Englishman can be arranged themselves in 4! Ways.
Hence, required number of ways = 3! 8! 4! 4! Ways.
Solution:
Taking all person of same nationality as one person, then we will have only three people.These three person can be arranged themselves in 3! Ways.
8 Indians can be arranged themselves in 8! Way.
4 American can be arranged themselves in 4! Ways.
4 Englishman can be arranged themselves in 4! Ways.
Hence, required number of ways = 3! 8! 4! 4! Ways.
2. How many Permutations of the letters of the word APPLE are there?
But two letters PP is of same kind.
Thus, required permutations,
Solution:
APPLE = 5 letters. But two letters PP is of same kind.
Thus, required permutations,
3. How many different words can be formed using all the letters of the word ALLAHABAD?
(a) When vowels occupy the even positions.
(b) Both L do not occur together.
So, permutations = = 7560
(a) There are 4 vowels and all are alike i.e. 4A's.
_2nd _4th _6th _8th _
These even places can be occupied by 4 vowels. In = 1 Way.
In other five places 5 other letter can be occupied of which two are alike i.e. 2L's.
Number of ways = Ways.
Hence, total number of ways in which vowels occupy the even places = × 1 = 60 ways.
(b) Taking both L's together and treating them as one letter we have 8 letters out of which A repeats 4 times and others are distinct.
These 8 letters can be arranged in = 1680 ways.
Also two L can be arranged themselves in 2! ways.
So, Total no. of ways in which L are together = 1680 × 2 = 3360 ways.
Now,
Total arrangement in which L never occur together,
= Total arrangement - Total no. of ways in which L occur together.
= 7560 - 3360
= 4200 ways
(a) When vowels occupy the even positions.
(b) Both L do not occur together.
Solution:
ALLAHABAD = 9 letters. Out of these 9 letters there is 4 A's and 2 L's are there.So, permutations = = 7560
(a) There are 4 vowels and all are alike i.e. 4A's.
_2nd _4th _6th _8th _
These even places can be occupied by 4 vowels. In = 1 Way.
In other five places 5 other letter can be occupied of which two are alike i.e. 2L's.
Number of ways = Ways.
Hence, total number of ways in which vowels occupy the even places = × 1 = 60 ways.
(b) Taking both L's together and treating them as one letter we have 8 letters out of which A repeats 4 times and others are distinct.
These 8 letters can be arranged in = 1680 ways.
Also two L can be arranged themselves in 2! ways.
So, Total no. of ways in which L are together = 1680 × 2 = 3360 ways.
Now,
Total arrangement in which L never occur together,
= Total arrangement - Total no. of ways in which L occur together.
= 7560 - 3360
= 4200 ways
4. In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10! - 9! × 2!
= 9!(10 - 2)
= 8 × 9!
Solution:
No. of ways in which 10 paper can arranged is 10! Ways.When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10! - 9! × 2!
= 9!(10 - 2)
= 8 × 9!
5. In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.
B G B G B G B
4 boys can be seated in 4! Ways
Girl can be seated in 3! Ways
Required number of ways,
= 4! × 3!
= 144
Solution:
Let the Arrangement be,B G B G B G B
4 boys can be seated in 4! Ways
Girl can be seated in 3! Ways
Required number of ways,
= 4! × 3!
= 144
6. A two member committee comprising of one male and one female member is to be constitute out of five males and three females. Amongst the females. Ms. A refuses to be a member of the committee in which Mr. B is taken as the member. In how many different ways can the committee be constituted ?
= 15 - 1
= 14
Solution:
5C1 × 3C1 - 1= 15 - 1
= 14
7. In how many ways 2 students can be chosen from the class of 20 students?
Solution:
Number of ways
8. Three gentlemen and three ladies are candidates for two vacancies. A voter has to vote for two candidates. In how many ways can one cast his vote?
So, the required number of ways is,
Solution:
There are 6 candidates and a voter has to vote for any two of them.So, the required number of ways is,
9. A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions?
Similarly, 5 questions can be chosen from 10 questions of Part B as = 10C5
Hence, total number of ways,
Solution:
There 10 questions in part A out of which 8 question can be chosen as = 10C8Similarly, 5 questions can be chosen from 10 questions of Part B as = 10C5
Hence, total number of ways,
10. Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices.
And polygon of 8 sides has 8 angular points.
Hence, number of triangle formed,
Solution:
A triangle needs 3 points.And polygon of 8 sides has 8 angular points.
Hence, number of triangle formed,
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