Anonymous
6/13/2025, 9:44:04 PM No.16697095
read the explanation for this in the book but didn't really get it. help me get the intuition for this /sci/
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Thread 16697095 - /sci/ [Archived: 1029 hours ago]
Anonymous
6/13/2025, 10:21:19 PM No.16697120
>>16697095 (OP)
It's the coefficient of x^18 in
(1 + x +... + x^7)(1 + x + ... + x^8) (1 + x + .. + x^9).
Now do algebraic manipulations.
It's the coefficient of x^18 in
(1 + x +... + x^7)(1 + x + ... + x^8) (1 + x + .. + x^9).
Now do algebraic manipulations.
Anonymous
6/14/2025, 1:42:59 PM No.16697527
boomp
explain scicucks
explain scicucks
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Anonymous
6/14/2025, 8:07:21 PM No.16697794
>>16697095 (OP)
Can't believe someone hasn't solved this yet.
I'm going to assume order doesn't matter here.
The problem is essentially equivalent to "how many ordered triples of integers [math](a,b,c)[/math] satisfy [math](a\leq 7) \land (b\leq 8) \land (c\leq 9) \land (a+b+c=18)[/math]?"
The last condition can be used to simplify the problem [math](c=18-a-b) \Longrightarrow (18-a-b \leq 9) \Longrightarrow (a+b \geq 9)[/math]
So the problem reduces to "how many ordered pairs of integers [math](a,b)[/math] satisfy [math](a\leq 7) \land (b\leq 8) \land (a+b \geq 9)[/math]?"
Notice that [math]a[/math] can't be 0, so if we start with [math]a=1[/math], then [math]b[/math] can take only one value, namely 8. If [math]a=2[/math] then [math]b[/math] can take on two values, namely 8 and 7, etc.. and generally if [math]a=x[/math] then [math]b[/math] can take on [math]x[/math] values... so the number of such ordered pairs is[eqn]\sum_{i=1}^{7}i =28[/eqn]I'm sure there is another simpler way to do it but I'm too lazy to think more about this problem.
Can't believe someone hasn't solved this yet.
I'm going to assume order doesn't matter here.
The problem is essentially equivalent to "how many ordered triples of integers [math](a,b,c)[/math] satisfy [math](a\leq 7) \land (b\leq 8) \land (c\leq 9) \land (a+b+c=18)[/math]?"
The last condition can be used to simplify the problem [math](c=18-a-b) \Longrightarrow (18-a-b \leq 9) \Longrightarrow (a+b \geq 9)[/math]
So the problem reduces to "how many ordered pairs of integers [math](a,b)[/math] satisfy [math](a\leq 7) \land (b\leq 8) \land (a+b \geq 9)[/math]?"
Notice that [math]a[/math] can't be 0, so if we start with [math]a=1[/math], then [math]b[/math] can take only one value, namely 8. If [math]a=2[/math] then [math]b[/math] can take on two values, namely 8 and 7, etc.. and generally if [math]a=x[/math] then [math]b[/math] can take on [math]x[/math] values... so the number of such ordered pairs is[eqn]\sum_{i=1}^{7}i =28[/eqn]I'm sure there is another simpler way to do it but I'm too lazy to think more about this problem.
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Anonymous
6/16/2025, 7:03:19 PM No.16699416
Anonymous
6/16/2025, 7:12:47 PM No.16699424
>>16697095 (OP)
Count it all up. For example.
C(7,2) + C(8,7) + C(9,9)
This is 2A and 7B and 9C. Then add up all the possibilities. It's not a terribly complicated sum.
Count it all up. For example.
C(7,2) + C(8,7) + C(9,9)
This is 2A and 7B and 9C. Then add up all the possibilities. It's not a terribly complicated sum.
Anonymous
6/17/2025, 3:52:00 AM No.16699668
>>16697095 (OP)
7+8+9=24. 24-18=6. That means there's 6 letters that you don't pick. So you want to distribute 6 identical balls into 3 non identical boxes, or 6 Scarlett letters onto 3 bitches, however you like. How many ways are there of doing this? Stars and bars. Specifically, 6 stars and 2 bars. **|**|** Is an example of this, representing the scenario with each bitch getting 2 Scarlett letters. Now use the stars and bars formula: (6+3-1)C(3-1) =8C2=8!/(2!6!)= 8*7/2=28. If I've explained to a satisfactory level, please delete the thread
7+8+9=24. 24-18=6. That means there's 6 letters that you don't pick. So you want to distribute 6 identical balls into 3 non identical boxes, or 6 Scarlett letters onto 3 bitches, however you like. How many ways are there of doing this? Stars and bars. Specifically, 6 stars and 2 bars. **|**|** Is an example of this, representing the scenario with each bitch getting 2 Scarlett letters. Now use the stars and bars formula: (6+3-1)C(3-1) =8C2=8!/(2!6!)= 8*7/2=28. If I've explained to a satisfactory level, please delete the thread
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Anonymous
6/17/2025, 4:48:32 AM No.16699709
>>16697095 (OP)
it’s not even clear from your screenshot whether order matters or not. since you have the book, can you clarify?
it’s not even clear from your screenshot whether order matters or not. since you have the book, can you clarify?
Anonymous
6/17/2025, 8:38:27 AM No.16699804
>>16699668
Correct answer. The rest of the thread is retards, as usual.
Correct answer. The rest of the thread is retards, as usual.