Minimum number of combinations under certain conditions
$begingroup$
A tutorial sheet has n problems. My answer sheet must contain solutions to all problems, and in order to pass I have to solve at least r problems correctly. I solve all n problems, but I do not know which ones I solved correctly…
I ask my tutor for help and he tells me that I have solved exactly r problems correctly, without specifying which ones. Therefore, there can be a minimum of nCr = n! / r!(n - r)! answer sheets, out of which only one contains r correct solutions.
After spending some more time on my solutions, I ask again my tutor for help and he tells me that I now have solved exactly r +1 problems correctly, again without specifying which ones.
My question is: what is the minimum number, x, of answer sheets, out of which at least one contains exactly r correct solutions, given r + m (m = 1, 2, 3, …) correct solutions in total (r + m <= n – 1)?
I have worked out a few cases manually, without finding an expression… for example:
- n = 6, r = 3, m = 1, then x = 6 answer sheets, out of which 1-2 have exactly r = 3 correct solutions (depending of the choices)
- n = 7, r = 4, m = 1, then x = 7 answer sheets, out of which only 1 has exactly r = 4 correct solutions
- n = 8, r = 2, m = 1, then x = 12 answer sheets, out of which 1-3 have exactly r = 2 correct solutions (depending of the choices)
- n = 8, r = 2, m = 2, then x = 7 answer sheets, out of which 1-3 have exactly r = 2 correct solutions (depending of the choices)
Can somebody please help me find an expression for this?
combinatorics combinations
asked Dec 24 '18 at 20:49
MakisMakis
12
$endgroup$
add a comment |
$begingroup$
A tutorial sheet has n problems. My answer sheet must contain solutions to all problems, and in order to pass I have to solve at least r problems correctly. I solve all n problems, but I do not know which ones I solved correctly…
I ask my tutor for help and he tells me that I have solved exactly r problems correctly, without specifying which ones. Therefore, there can be a minimum of nCr = n! / r!(n - r)! answer sheets, out of which only one contains r correct solutions.
After spending some more time on my solutions, I ask again my tutor for help and he tells me that I now have solved exactly r +1 problems correctly, again without specifying which ones.
My question is: what is the minimum number, x, of answer sheets, out of which at least one contains exactly r correct solutions, given r + m (m = 1, 2, 3, …) correct solutions in total (r + m <= n – 1)?
I have worked out a few cases manually, without finding an expression… for example:
- n = 6, r = 3, m = 1, then x = 6 answer sheets, out of which 1-2 have exactly r = 3 correct solutions (depending of the choices)
- n = 7, r = 4, m = 1, then x = 7 answer sheets, out of which only 1 has exactly r = 4 correct solutions
- n = 8, r = 2, m = 1, then x = 12 answer sheets, out of which 1-3 have exactly r = 2 correct solutions (depending of the choices)
- n = 8, r = 2, m = 2, then x = 7 answer sheets, out of which 1-3 have exactly r = 2 correct solutions (depending of the choices)
Can somebody please help me find an expression for this?
combinatorics combinations
asked Dec 24 '18 at 20:49
MakisMakis
12
$endgroup$
add a comment |
$begingroup$
A tutorial sheet has n problems. My answer sheet must contain solutions to all problems, and in order to pass I have to solve at least r problems correctly. I solve all n problems, but I do not know which ones I solved correctly…
I ask my tutor for help and he tells me that I have solved exactly r problems correctly, without specifying which ones. Therefore, there can be a minimum of nCr = n! / r!(n - r)! answer sheets, out of which only one contains r correct solutions.
After spending some more time on my solutions, I ask again my tutor for help and he tells me that I now have solved exactly r +1 problems correctly, again without specifying which ones.
My question is: what is the minimum number, x, of answer sheets, out of which at least one contains exactly r correct solutions, given r + m (m = 1, 2, 3, …) correct solutions in total (r + m <= n – 1)?
I have worked out a few cases manually, without finding an expression… for example:
- n = 6, r = 3, m = 1, then x = 6 answer sheets, out of which 1-2 have exactly r = 3 correct solutions (depending of the choices)
- n = 7, r = 4, m = 1, then x = 7 answer sheets, out of which only 1 has exactly r = 4 correct solutions
- n = 8, r = 2, m = 1, then x = 12 answer sheets, out of which 1-3 have exactly r = 2 correct solutions (depending of the choices)
- n = 8, r = 2, m = 2, then x = 7 answer sheets, out of which 1-3 have exactly r = 2 correct solutions (depending of the choices)
Can somebody please help me find an expression for this?
combinatorics combinations
asked Dec 24 '18 at 20:49
MakisMakis
12
$endgroup$
A tutorial sheet has n problems. My answer sheet must contain solutions to all problems, and in order to pass I have to solve at least r problems correctly. I solve all n problems, but I do not know which ones I solved correctly…
I ask my tutor for help and he tells me that I have solved exactly r problems correctly, without specifying which ones. Therefore, there can be a minimum of nCr = n! / r!(n - r)! answer sheets, out of which only one contains r correct solutions.
After spending some more time on my solutions, I ask again my tutor for help and he tells me that I now have solved exactly r +1 problems correctly, again without specifying which ones.
My question is: what is the minimum number, x, of answer sheets, out of which at least one contains exactly r correct solutions, given r + m (m = 1, 2, 3, …) correct solutions in total (r + m <= n – 1)?
I have worked out a few cases manually, without finding an expression… for example:
- n = 6, r = 3, m = 1, then x = 6 answer sheets, out of which 1-2 have exactly r = 3 correct solutions (depending of the choices)
- n = 7, r = 4, m = 1, then x = 7 answer sheets, out of which only 1 has exactly r = 4 correct solutions
- n = 8, r = 2, m = 1, then x = 12 answer sheets, out of which 1-3 have exactly r = 2 correct solutions (depending of the choices)
- n = 8, r = 2, m = 2, then x = 7 answer sheets, out of which 1-3 have exactly r = 2 correct solutions (depending of the choices)
Can somebody please help me find an expression for this?
combinatorics combinations
combinatorics combinations
asked Dec 24 '18 at 20:49
MakisMakis
12
asked Dec 24 '18 at 20:49
MakisMakis
12
asked Dec 24 '18 at 20:49
MakisMakis
12
asked Dec 24 '18 at 20:49
MakisMakis
12
asked Dec 24 '18 at 20:49
MakisMakis
12
12
add a comment |
add a comment |
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