Discrete Math: Combination with Repetitions
QUESTION: During a period of 7 days, Charles eats a total of
25 donuts. A donut schedule is a sequence of 7 numbers, whose sum is equal to 25, and whose numbers indicate the number of donuts that Charles eats on each day. Three examples of such schedules are (3; 2; 7; 4; 1; 3; 5), (2; 3; 7; 4; 1; 3; 5), and (3; 0; 9; 4; 1; 0; 8). How many donut schedules are there?
Answer: 31C6
ATTEMPT: I realized that this is a combination with repetition problem. I used the formula (r+n-1)! / r! * (n-1)! where r is the number of slots and n is the number of options. I used 7 as the number of slots and 25 as the number of options to get my answer to be 31C7. Don't know what I have missed accounting here.
combinatorics
asked Nov 25 at 5:24
Toby
1577
add a comment |
QUESTION: During a period of 7 days, Charles eats a total of
25 donuts. A donut schedule is a sequence of 7 numbers, whose sum is equal to 25, and whose numbers indicate the number of donuts that Charles eats on each day. Three examples of such schedules are (3; 2; 7; 4; 1; 3; 5), (2; 3; 7; 4; 1; 3; 5), and (3; 0; 9; 4; 1; 0; 8). How many donut schedules are there?
Answer: 31C6
ATTEMPT: I realized that this is a combination with repetition problem. I used the formula (r+n-1)! / r! * (n-1)! where r is the number of slots and n is the number of options. I used 7 as the number of slots and 25 as the number of options to get my answer to be 31C7. Don't know what I have missed accounting here.
combinatorics
asked Nov 25 at 5:24
Toby
1577
add a comment |
QUESTION: During a period of 7 days, Charles eats a total of
25 donuts. A donut schedule is a sequence of 7 numbers, whose sum is equal to 25, and whose numbers indicate the number of donuts that Charles eats on each day. Three examples of such schedules are (3; 2; 7; 4; 1; 3; 5), (2; 3; 7; 4; 1; 3; 5), and (3; 0; 9; 4; 1; 0; 8). How many donut schedules are there?
Answer: 31C6
ATTEMPT: I realized that this is a combination with repetition problem. I used the formula (r+n-1)! / r! * (n-1)! where r is the number of slots and n is the number of options. I used 7 as the number of slots and 25 as the number of options to get my answer to be 31C7. Don't know what I have missed accounting here.
combinatorics
asked Nov 25 at 5:24
Toby
1577
QUESTION: During a period of 7 days, Charles eats a total of
25 donuts. A donut schedule is a sequence of 7 numbers, whose sum is equal to 25, and whose numbers indicate the number of donuts that Charles eats on each day. Three examples of such schedules are (3; 2; 7; 4; 1; 3; 5), (2; 3; 7; 4; 1; 3; 5), and (3; 0; 9; 4; 1; 0; 8). How many donut schedules are there?
Answer: 31C6
ATTEMPT: I realized that this is a combination with repetition problem. I used the formula (r+n-1)! / r! * (n-1)! where r is the number of slots and n is the number of options. I used 7 as the number of slots and 25 as the number of options to get my answer to be 31C7. Don't know what I have missed accounting here.
combinatorics
combinatorics
asked Nov 25 at 5:24
Toby
1577
asked Nov 25 at 5:24
Toby
1577
asked Nov 25 at 5:24
Toby
1577
asked Nov 25 at 5:24
Toby
1577
asked Nov 25 at 5:24
Toby
1577
1577
add a comment |
add a comment |
1 Answer
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Imagine you have $25$ donuts line up, then you want to make cuts to the line of donuts to divide the donuts into $7$ parts. You need $6$ cuts to divide a line into $7$ parts. So imagine $25+6$ position on a line where you can either place a cut or a donuts, you have to choose 6 out of the 31 positions to place the cuts. So that is $binom{25+6}{6}$ ways to do it. Is this correct?
So here is an example of how to relate a sequence of donuts and cuts to a schedule: Here I'll use D to denote donuts and c to denote cut.
ccdddddddddddddddddddddddddcccc correspond to $(0,0,25,0,0,0,0)$.
cddccddddddddddddddddddddddcdcc correspond to $(0,2,0,22,1,0,0)$.
So you see you are choosing 6 position to put cuts in 31 position and the rest to put donuts. 6 cuts because 6 cuts divide a line into 7 parts.
answered Nov 25 at 5:35
mathnoob
1,794422
Yes, the answer you got is correct. I am still confused on how you got the 6 position part, like the placing a cut or donut part. What part of the question should I be analyzing that I can draw the conclusion that even though it says 7 number sequence, by considering drawing lines it should actually be 6?
– Toby
Nov 25 at 5:49
add a comment |
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1 Answer
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Imagine you have $25$ donuts line up, then you want to make cuts to the line of donuts to divide the donuts into $7$ parts. You need $6$ cuts to divide a line into $7$ parts. So imagine $25+6$ position on a line where you can either place a cut or a donuts, you have to choose 6 out of the 31 positions to place the cuts. So that is $binom{25+6}{6}$ ways to do it. Is this correct?
So here is an example of how to relate a sequence of donuts and cuts to a schedule: Here I'll use D to denote donuts and c to denote cut.
ccdddddddddddddddddddddddddcccc correspond to $(0,0,25,0,0,0,0)$.
cddccddddddddddddddddddddddcdcc correspond to $(0,2,0,22,1,0,0)$.
So you see you are choosing 6 position to put cuts in 31 position and the rest to put donuts. 6 cuts because 6 cuts divide a line into 7 parts.
answered Nov 25 at 5:35
mathnoob
1,794422
Yes, the answer you got is correct. I am still confused on how you got the 6 position part, like the placing a cut or donut part. What part of the question should I be analyzing that I can draw the conclusion that even though it says 7 number sequence, by considering drawing lines it should actually be 6?
– Toby
Nov 25 at 5:49
add a comment |
Imagine you have $25$ donuts line up, then you want to make cuts to the line of donuts to divide the donuts into $7$ parts. You need $6$ cuts to divide a line into $7$ parts. So imagine $25+6$ position on a line where you can either place a cut or a donuts, you have to choose 6 out of the 31 positions to place the cuts. So that is $binom{25+6}{6}$ ways to do it. Is this correct?
So here is an example of how to relate a sequence of donuts and cuts to a schedule: Here I'll use D to denote donuts and c to denote cut.
ccdddddddddddddddddddddddddcccc correspond to $(0,0,25,0,0,0,0)$.
cddccddddddddddddddddddddddcdcc correspond to $(0,2,0,22,1,0,0)$.
So you see you are choosing 6 position to put cuts in 31 position and the rest to put donuts. 6 cuts because 6 cuts divide a line into 7 parts.
answered Nov 25 at 5:35
mathnoob
1,794422
Yes, the answer you got is correct. I am still confused on how you got the 6 position part, like the placing a cut or donut part. What part of the question should I be analyzing that I can draw the conclusion that even though it says 7 number sequence, by considering drawing lines it should actually be 6?
– Toby
Nov 25 at 5:49
add a comment |
Imagine you have $25$ donuts line up, then you want to make cuts to the line of donuts to divide the donuts into $7$ parts. You need $6$ cuts to divide a line into $7$ parts. So imagine $25+6$ position on a line where you can either place a cut or a donuts, you have to choose 6 out of the 31 positions to place the cuts. So that is $binom{25+6}{6}$ ways to do it. Is this correct?
So here is an example of how to relate a sequence of donuts and cuts to a schedule: Here I'll use D to denote donuts and c to denote cut.
ccdddddddddddddddddddddddddcccc correspond to $(0,0,25,0,0,0,0)$.
cddccddddddddddddddddddddddcdcc correspond to $(0,2,0,22,1,0,0)$.
So you see you are choosing 6 position to put cuts in 31 position and the rest to put donuts. 6 cuts because 6 cuts divide a line into 7 parts.
answered Nov 25 at 5:35
mathnoob
1,794422
Imagine you have $25$ donuts line up, then you want to make cuts to the line of donuts to divide the donuts into $7$ parts. You need $6$ cuts to divide a line into $7$ parts. So imagine $25+6$ position on a line where you can either place a cut or a donuts, you have to choose 6 out of the 31 positions to place the cuts. So that is $binom{25+6}{6}$ ways to do it. Is this correct?
So here is an example of how to relate a sequence of donuts and cuts to a schedule: Here I'll use D to denote donuts and c to denote cut.
ccdddddddddddddddddddddddddcccc correspond to $(0,0,25,0,0,0,0)$.
cddccddddddddddddddddddddddcdcc correspond to $(0,2,0,22,1,0,0)$.
So you see you are choosing 6 position to put cuts in 31 position and the rest to put donuts. 6 cuts because 6 cuts divide a line into 7 parts.
answered Nov 25 at 5:35
mathnoob
1,794422
edited Nov 25 at 6:07
answered Nov 25 at 5:35
mathnoob
1,794422
answered Nov 25 at 5:35
mathnoob
1,794422
answered Nov 25 at 5:35
mathnoob
1,794422
1,794422
Yes, the answer you got is correct. I am still confused on how you got the 6 position part, like the placing a cut or donut part. What part of the question should I be analyzing that I can draw the conclusion that even though it says 7 number sequence, by considering drawing lines it should actually be 6?
– Toby
Nov 25 at 5:49
add a comment |
Yes, the answer you got is correct. I am still confused on how you got the 6 position part, like the placing a cut or donut part. What part of the question should I be analyzing that I can draw the conclusion that even though it says 7 number sequence, by considering drawing lines it should actually be 6?
– Toby
Nov 25 at 5:49
Yes, the answer you got is correct. I am still confused on how you got the 6 position part, like the placing a cut or donut part. What part of the question should I be analyzing that I can draw the conclusion that even though it says 7 number sequence, by considering drawing lines it should actually be 6?
– Toby
Nov 25 at 5:49
Yes, the answer you got is correct. I am still confused on how you got the 6 position part, like the placing a cut or donut part. What part of the question should I be analyzing that I can draw the conclusion that even though it says 7 number sequence, by considering drawing lines it should actually be 6?
– Toby
Nov 25 at 5:49
add a comment |
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