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How many solutions are there to the equation
u + v + w + x + y + z = 18
where u, v, w, x, y, and z are nonnegative integers, and x is at most 10?

 Dec 14, 2022
 #1
avatar+24 
0

 

There are 656 solutions to this equation. This can be calculated using the stars and bars method. The equation can be rewritten as u + v + w + (x + y + z) = 18. This can be represented as a stars and bars problem, with 18 stars and 5 bars. The number of ways to arrange the stars and bars is equal to the number of solutions to the equation. The number of ways to arrange 18 stars and 5 bars is equal to the number of ways to choose 5 positions from 23 positions (18 stars + 5 bars). This can be calculated using the combination formula: C(23, 5) = 23! / (5! * 18!) = 23 * 22 * 21 * 20 * 19 / (5 * 4 * 3 * 2 * 1) = 23 * 22 * 21 * 20 * 19 / 120 = 656 Therefore, there are 656 solutions to the equation u + v + w + x + y + z = 18, where u, v, w, x, y, and z are nonnegative integers, and x is at most 10.

 Dec 14, 2022
 #2
avatar+2541 
0

By stars and bars, there are 5 bars and 18 stars, so there are \({23 \choose 5} = 33,649\)

 

But, this overcounts for the cases where x is greater than 10. 

 

If x is 18, there are 0 stars and 4 bars which makes for \({4 \choose 4} = 1 \) case. 

If x is 17, there are \({5 \choose 4} = 5\) cases

If x is 16, there are \({6 \choose 4} = 15\)

 

Notice a pattern?

 

This pattern will continue until \({11 \choose 4}\). Summing all these up gives us \(792\) cases.

 

So, there are \(33649 - 792 = \color{brown}\boxed{32,857} \) cases.

 Dec 14, 2022

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