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6/14/2025, 8:07:21 PM
>>16697095
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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