# How many solutions possible for the equation $x_1+x_2+x_3+x_4+x_5=55$ if

How several remedies feasible for the formula $x_1+x_2+x_3+x_4+x_5=55$ if all $x$ are non - adverse integer:

- No constraints. The remedy is $C(55 + 4, 4) = C(59,4)$ yet I fall short to see why, can a person clarify this to me?
- Every $x_k$ is weird
- If $x_1>=1,x_2>=2,x_3>=3,x_4>=2,x_5>=1,$

8

meiryo 2022-06-07 14:35:41

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Answers: 1

- You neglected to claim that $0\le x_i$ (or else there is a boundless variety of remedies). This is a typical "choice with reps and also without order" instance (therefore the formula, which is just one of the standard solutions in combinatorics). In this instance, you have 55 "balls" to disperse openly right into the 5 "holes" - the variables
- Here you can write $x_i=2y_i+1$ and also address for $y_i$ as opposed to $x_i$. You'll get something like $2(y_1+\dots+y_5)+5=55$ - streamline to get $y_1+\dots+y_5=25$ (just how?)
- Below you make use of $x_1=y_1+1$, $x_2=y_2+2$ and afterwards once more address for $y_i$

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Gadi A 2022-06-07 14:57:57

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