26 Jan 21
The puzzles in the Saturday Times have been rubbish.I thought this was the same but:-
"4 friends each order some chicken at a restaurant.They order 11 pieces between then.In how many ways could this be done?
Too easy says I.11!/4! = 330.
This is not the answer given.
Hint:- Nor is the answer permutations which would be higher
What is the answer and what am I doing wrong?
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26 Jan 21
@venda saidI believe they are looking for what is called partitions of 11 ( i.e. how many ways can you make the sum 11 with 4 or less summands ) or something along those lines.
The puzzles in the Saturday Times have been rubbish.I thought this was the same but:-
"4 friends each order some chicken at a restaurant.They order 11 pieces between then.In how many ways could this be done?
Too easy says I.11!/4! = 330.
This is not the answer given.
Hint:- Nor is the answer permutations which would be higher
What is the answer and what am I doing wrong?
26 Jan 21
@joe-shmo saidThis is interesting stuff.I'd never heard of partition theory but it seems to be very complicated.Apparently there is no simple formula such as (n!-r!)/(n-r)! * r!) which we all know but one needs to write algorithms or use diagrams.
I believe they are looking for what is called partitions of 11 ( i.e. how many ways can you make the sum 11 with 4 or less summands ) or something along those lines.
This is beyond most casual Times readers I would have thought.
Surely they aren't expecting readers to work it out manually? e.g 8+1+1+1 , 7+2+1+1 and so on
26 Jan 21
@joe-shmo saidActually they look for the sum of eleven with four summands:
I believe they are looking for what is called partitions of 11 ( i.e. how many ways can you make the sum 11 with 4 or less summands ) or something along those lines.
8+1+1+1
7+2+1+1
6+3+1+1 6+2+2+1
5+4+1+1 5+3+2+1 5+2+2+2
4+4+2+1 4+3+3+1 4+3+2+2
3+3+3+2
This should be the solution (11 cases)
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26 Jan 21
1 edit
@ponderable saidI think you have to add partitions of 11 with 3, 2, 1 summands into that group as well ( unless they stipulate that everyone must but at least one piece). And then I would think we have to multiply by 4
Actually they look for the sum of eleven with four summands:
8+1+1+1
7+2+1+1
6+3+1+1 6+2+2+1
5+4+1+1 5+3+2+1 5+2+2+2
4+4+2+1 4+3+3+1 4+3+2+2
3+3+3+2
This should be the solution (11 cases)
If we have person A,B,C,D for just a single person getting all 11
( 11,0,0,0,) A gets them all
( 0,11,0,0) B gets them all
etc...
26 Jan 21
@joe-shmo saidThat is implicit in "each orders" I think
I think you have to add partitions of 11 with 3, 2, 1 summands as well ( unless they stipulate that everyone must but at least one piece.
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26 Jan 21
1 edit
@ponderable saidWell, we still should multiply by 4 as each partition has 4 distinct orderings?
That is implicit in "each orders" I think
26 Jan 21
@joe-shmo saidYou can do that. Though since the people were not identified in the first place I think the answer they looked for is "just" the different cases.
Well, we still should multiply by 4 as each partition has 4 distinct orderings?
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26 Jan 21
5 edits
@ponderable saidI feel like this is missing many cases?
Actually they look for the sum of eleven with four summands:
8+1+1+1
7+2+1+1
6+3+1+1 6+2+2+1
5+4+1+1 5+3+2+1 5+2+2+2
4+4+2+1 4+3+3+1 4+3+2+2
3+3+3+2
This should be the solution (11 cases)
persons A,B,C,D.
Examine cases where A & B each buy 1: We have 8 distinct orders here.
A B C D
1 + 1 +1 + 8
1 + 1 +2 + 7
1 + 1 +3 + 6
1 + 1 +4 + 5
1 + 1 +5 + 4
1 + 1 +6 + 3
1 + 1 +7 + 2
1 + 1 +8 + 1
Start to generalize:
Examine the case of 5 into 3 summands:
1 ○ 1 ○ 1 ○ 1 ○ 1
Fill the ○'s with + signs to form the sum:
1 + 1 + 1 ○ 1 ○ 1 = 1+1+3
1 + 1 ○ 1 + 1 ○ 1 = 1+2+2
1 + 1 ○ 1 ○ 1 + 1 = 1+3+1
1 ○ 1 + 1 + 1 ○ 1 = 2+1+2
1 ○ 1 + 1 ○ 1 + 1 = 2+2+1
1 ○ 1 ○ 1 + 1 + 1 = 3+1+1
So in general:
Let S = the Sum
N = number of Summands
What this amounts to is the number of combinations "C" we can fill (S-1) ○'s with (N-1) + signs
C(n,r) = n!/( r!*(n-r)! ) ( n objects choose r )
For this:
n = S-1
r = N-1
C = ( S -1 )!/ ( ( N-1)! ( S - N )! )
For my example we see S = 5, n = 3.
4! / ( 2! * 2! ) = 6
This matches the enumeration above.
For another example lets do Sum of 4 into 2 summands:
1 + 1 ○ 1 ○ 1 = 1 + 3
1 ○ 1 + 1 ○ 1 = 2 + 2
1 ○ 1 ○ 1 + 1 = 3 +1
The formula yeilds: S = 4, n = 2
C = 3!/( 1! * 2! ) = 3
Same as enumerated above
Extending this to the problem:
S = 11
n = 4
C = ( 11 -1 )!/ ( ( 4-1)! ( 11 - 4 )! ) = 10!/( 3! * 7! ) = 120 Ways
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26 Jan 21
1 edit
@venda saidYeah, the solution doesn't involve partitions exactly ( sorry for misleading you a bit ), just something related to partitions. Luckily this "something" ( the number ways to write the sum S of N summands ) we can derive a formula for ( as shown in my last post ).
This is interesting stuff.I'd never heard of partition theory but it seems to be very complicated.Apparently there is no simple formula such as (n!-r!)/(n-r)! * r!) which we all know but one needs to write algorithms or use diagrams.
This is beyond most casual Times readers I would have thought.
Surely they aren't expecting readers to work it out manually? e.g 8+1+1+1 , 7+2+1+1 and so on
26 Jan 21
@joe-shmo saidCorrect Joe!!
I feel like this is missing many cases?
persons A,B,C,D.
Examine cases where A & B each buy 1: We have 8 distinct orders here.
A B C D
1 + 1 +1 + 8
1 + 1 +2 + 7
1 + 1 +3 + 6
1 + 1 +4 + 5
1 + 1 +5 + 4
1 + 1 +6 + 3
1 + 1 +7 + 2
1 + 1 +8 + 1
Start to generalize:
Examine the case of 5 into 3 summands:
1 ○ 1 ○ 1 ○ 1 ○ 1
Fill the ○'s with + sig ...[text shortened]... problem:
S = 11
n = 4
C = ( 11 -1 )!/ ( ( 4-1)! ( 11 - 4 )! ) = 10!/( 3! * 7! ) = 120 Ways
Thanks .
I'll transcribe your formula into a spreadsheet for future reference.
These things fascinate me
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26 Jan 21
1 edit
@venda saidThanks Venda!
Correct Joe!!
Thanks .
I'll transcribe your formula into a spreadsheet for future reference.
These things fascinate me
It seems like they expect a little bit more mathematical ability from the average "joe shmo" reader from across the pond.
I assume the solution technique made good sense to you?
27 Jan 21
@joe-shmo saidI haven't studied it fully yet,but it seemed quite logical.
Thanks Venda!
It seems like they expect a little bit more mathematical ability from the average "joe shmo" reader from across the pond.
I assume the solution technique made good sense to you?
Thanks again.
As to what they expect,there's always a "listener" crossword in there as well.
I can't even work out how to fill it in ,never mind solve the clues but someone wins the prize every week!!
Suppose I'll have to accept that at 68 yrs old I'm not academically gifted and it's not likely to get much better!!
05 Feb 21
@venda saidMaybe you have to consider who gets how many wings, thighs, legs, breasts
The puzzles in the Saturday Times have been rubbish.I thought this was the same but:-
"4 friends each order some chicken at a restaurant.They order 11 pieces between then.In how many ways could this be done?
Too easy says I.11!/4! = 330.
This is not the answer given.
Hint:- Nor is the answer permutations which would be higher
What is the answer and what am I doing wrong?
- Joined
- 10 Dec 06
- Moves
- 8528
05 Feb 21
2 edits
@athousandyoung saidIf you read further, you'll see that they weren't looking for that.
Maybe you have to consider who gets how many wings, thighs, legs, breasts
BUT...that is a good question!
Bonus
If 4 friends order 11 pieces of chicken( each ordering at least one piece ) that can be of the type wing, leg, breast or thigh in how many ways can this be done?