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In an examination of 9 papers a candidate has to pass in more papers than the number of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful is
- a)255
- b)256
- c)193
- d)319
Correct answer is option 'B'. Can you explain this answer?
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In an examination of 9 papers a candidate has to pass in more papers t...
The candidate is unsuccessful if he fails in 9 or 8 or 7 or 6 or 5 papers.
∴ the number of ways to be unsuccessful
∴ the number of ways to be unsuccessful
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In an examination of 9 papers a candidate has to pass in more papers t...
Problem Analysis:
The candidate needs to pass more papers than the number of papers in which he fails in order to be successful. Let's assume that the candidate passes x papers and fails y papers. According to the given condition, x > y.
Solution:
To find the number of ways in which the candidate can be unsuccessful, we need to consider all possible values of y and calculate the corresponding values of x.
Case 1: y = 0
If the candidate fails in 0 papers, it means he passes in all 9 papers. However, according to the given condition, he needs to pass more papers than the number of papers in which he fails. Therefore, this case is not valid.
Case 2: y = 1
If the candidate fails in 1 paper, he can pass in any number of papers from 2 to 9. The number of ways in which he can pass in 2 papers is 1, the number of ways in which he can pass in 3 papers is 8C3 = 56, and so on. The total number of ways in which he can be unsuccessful in this case is the sum of these values, which is given by:
1 + 8C3 + 8C4 + 8C5 + 8C6 + 8C7 + 8C8 + 8C9 = 1 + 56 + 70 + 56 + 28 + 8 + 1 = 220
Case 3: y = 2
If the candidate fails in 2 papers, he can pass in any number of papers from 3 to 9. Similarly, we can calculate the number of ways in which he can be unsuccessful in this case:
1 + 7C3 + 7C4 + 7C5 + 7C6 + 7C7 + 7C8 + 7C9 = 1 + 35 + 35 + 21 + 7 + 1 = 100
Case 4: y = 3
Similarly, if the candidate fails in 3 papers, he can pass in any number of papers from 4 to 9. We can calculate the number of ways in which he can be unsuccessful in this case:
1 + 6C3 + 6C4 + 6C5 + 6C6 + 6C7 + 6C8 + 6C9 = 1 + 20 + 15 + 6 + 1 = 43
Case 5: y = 4
1 + 5C3 + 5C4 + 5C5 + 5C6 + 5C7 + 5C8 + 5C9 = 1 + 10 + 5 + 1 = 17
Case 6: y = 5
1 + 4C3 + 4C4 + 4C5 + 4C6 + 4C7 + 4C8 + 4C9 = 1 + 4 + 1 = 6
Case 7: y = 6
1 + 3C3 + 3C4 + 3
The candidate needs to pass more papers than the number of papers in which he fails in order to be successful. Let's assume that the candidate passes x papers and fails y papers. According to the given condition, x > y.
Solution:
To find the number of ways in which the candidate can be unsuccessful, we need to consider all possible values of y and calculate the corresponding values of x.
Case 1: y = 0
If the candidate fails in 0 papers, it means he passes in all 9 papers. However, according to the given condition, he needs to pass more papers than the number of papers in which he fails. Therefore, this case is not valid.
Case 2: y = 1
If the candidate fails in 1 paper, he can pass in any number of papers from 2 to 9. The number of ways in which he can pass in 2 papers is 1, the number of ways in which he can pass in 3 papers is 8C3 = 56, and so on. The total number of ways in which he can be unsuccessful in this case is the sum of these values, which is given by:
1 + 8C3 + 8C4 + 8C5 + 8C6 + 8C7 + 8C8 + 8C9 = 1 + 56 + 70 + 56 + 28 + 8 + 1 = 220
Case 3: y = 2
If the candidate fails in 2 papers, he can pass in any number of papers from 3 to 9. Similarly, we can calculate the number of ways in which he can be unsuccessful in this case:
1 + 7C3 + 7C4 + 7C5 + 7C6 + 7C7 + 7C8 + 7C9 = 1 + 35 + 35 + 21 + 7 + 1 = 100
Case 4: y = 3
Similarly, if the candidate fails in 3 papers, he can pass in any number of papers from 4 to 9. We can calculate the number of ways in which he can be unsuccessful in this case:
1 + 6C3 + 6C4 + 6C5 + 6C6 + 6C7 + 6C8 + 6C9 = 1 + 20 + 15 + 6 + 1 = 43
Case 5: y = 4
1 + 5C3 + 5C4 + 5C5 + 5C6 + 5C7 + 5C8 + 5C9 = 1 + 10 + 5 + 1 = 17
Case 6: y = 5
1 + 4C3 + 4C4 + 4C5 + 4C6 + 4C7 + 4C8 + 4C9 = 1 + 4 + 1 = 6
Case 7: y = 6
1 + 3C3 + 3C4 + 3
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In an examination of 9 papers a candidate has to pass in more papers than the number of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful isa)255b)256c)193d)319Correct answer is option 'B'. Can you explain this answer?
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