# Combinatorics question with unequality and different subscript a) x 1 </msub>

Combinatorics question with unequality and different subscript
a) ${x}_{1}+{x}_{2}+...+{x}_{7}\le 30$ where ${x}_{i}^{\prime }s$ are even non-negative integers.
b) ${x}_{1}+{x}_{2}+...+{x}_{7}\le 30$ where ${x}_{i}^{\prime }s$ are odd non-negative integers.
c) ${x}_{1}+{x}_{2}+...+{x}_{6}\le 30$ where ${x}_{i}^{\prime }s$ are odd non-negative integers.
These questions is from my textbook. I know to solve similar questions such that ${x}_{1}+{x}_{2}+...+{x}_{k}\le n$ where ${x}_{i}^{\prime }s$ are non-negative integers. We add an extra term on the lefthandside and the rest found by combination with repetition such that $\left(\genfrac{}{}{0}{}{n+\left(k+1\right)-1}{n}\right)$. However , what happens when they are even or odd , is there any special technique ? Moreover , as you see the part a and b differ from only subscripts , i think that there must be a reason behind different subscript in same question.Is there any reason ? Thanks in advance..
You can still ask an expert for help

## Want to know more about Discrete math?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Kamora Greer
Step 1
Finding the number of solutions to ${x}_{1}+{x}_{2}+\cdots +{x}_{7}\le 30$ where each of the ${x}_{i}$'s are even non-negative integers...
Rather than writing them as ${x}_{1},{x}_{2},\dots$ since they are even we can write them as $2{y}_{1},2{y}_{2},\dots$
Step 2
We have then $2{y}_{1}+2{y}_{2}+\cdots +2{y}_{7}\le 30$. Divide both sides by 2 and we are left with ${y}_{1}+{y}_{2}+\cdots +{y}_{7}\le 15$ where each of the ${y}_{i}$'s are non-negative integers with no restrictions on parity.
Similarly, for the later problems where you deal with odd numbers, you can rewrite as $2{z}_{1}+1,2{z}_{2}+1,\dots$ You can then subtract the 1's from both sides and then use the same idea. The punchline is to rewrite a problem you haven't seen before in such a way that it becomes a problem you have seen before.
###### Not exactly what you’re looking for?
Brunton39
Step 1
For (a), you should solve for , where ${x}_{i}=2{y}_{i},{y}_{\ge }0$
For (b), you should solve for ${y}_{1}+{y}_{2}+...+{y}_{7}\le 11$ where ${x}_{i}=2{y}_{i}+1,{y}_{i}\ge 0$
For (c), you should solve for ${y}_{1}+{y}_{2}+...+{y}_{6}\le 12$ where ${x}_{i}=2{y}_{i}+1,{y}_{i}\ge 0$
Step 2
The difference between (b) and (c) is that in (b), there are 7 odd numbers and 7 odd numbers cannot sum to an even number so equality $\left(=30\right)$ is not possible and you would have sum $\le 29$. But in (c), you have 6 odd numbers so equality can be reached.