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

migongoniwt 2022-06-08 Answered
Combinatorics question with unequality and different subscript
a) x 1 + x 2 + . . . + x 7 30 where x i s are even non-negative integers.
b) x 1 + x 2 + . . . + x 7 30 where x i s are odd non-negative integers.
c) x 1 + x 2 + . . . + x 6 30 where x i 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 n where x i s are non-negative integers. We add an extra term on the lefthandside and the rest found by combination with repetition such that ( n + ( k + 1 ) 1 n ) . 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..
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Answers (2)

Kamora Greer
Answered 2022-06-09 Author has 16 answers
Step 1
Finding the number of solutions to x 1 + x 2 + + x 7 30 where each of the x i 's are even non-negative integers...
Rather than writing them as x 1 , x 2 , since they are even we can write them as 2 y 1 , 2 y 2 ,
Step 2
We have then 2 y 1 + 2 y 2 + + 2 y 7 30. Divide both sides by 2 and we are left with y 1 + y 2 + + y 7 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 , 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.
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Brunton39
Answered 2022-06-10 Author has 8 answers
Step 1
For (a), you should solve for y 1 + y 2 + . . . + y 7 15 ,  , where x i = 2 y i , y 0
For (b), you should solve for y 1 + y 2 + . . . + y 7 11 where x i = 2 y i + 1 , y i 0
For (c), you should solve for y 1 + y 2 + . . . + y 6 12 where x i = 2 y i + 1 , y i 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 ( = 30 ) is not possible and you would have sum 29. But in (c), you have 6 odd numbers so equality can be reached.
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