Can You Solve These Combinatorial Equations?

[IHateFactorial]

*sigh* As the title says, they are the bane of my existence... I'd really appreciate it if you guys could help me with these bloody things.

1. Prove that ₃C₀ + ₃C₁ + ₃C₂ + ₃C₃ = 2³ Generalize the formula for any value of r and n such that 0<=r<=n.

2. Prove that _n-1C_r + _n-1C_r-1 = _nC_r

3. i) How many solutions (in non-negative integers) are there of the equation x + y + z = 8?

3. ii) How many solutions (in non-negative integers) are there of the equation x + y + z = 18?

 

[Prove It]

*sigh* As the title says, they are the bane of my existence... I'd really appreciate it if you guys could help me with these bloody things.

1. Prove that ₃C₀ + ₃C₁ + ₃C₂ + ₃C₃ = 2³ Generalize the formula for any value of r and n such that 0<=r<=n.

2. Prove that _n-1C_r + _n-1C_r-1 = _nC_r

3. i) How many solutions (in non-negative integers) are there of the equation x + y + z = 8?

3. ii) How many solutions (in non-negative integers) are there of the equation x + y + z = 18?

Have you tried anything? The first two are just using the choose formula to simplify each...

 

[IHateFactorial]

Yes, I have. I do know that the first one simplifies to 1 + 3 + 3 + 1 which is 8, but the generalizing part stumps me.

As for the second, I have absolutely NO idea how to prove Pascal's Identity, aside from putting in the variables into their Combination formula.

Also, thanks for the sarcastic response, appreciate it.

 

[Prove It]

Yes, I have. I do know that the first one simplifies to 1 + 3 + 3 + 1 which is 8, but the generalizing part stumps me.

As for the second, I have absolutely NO idea how to prove Pascal's Identity, aside from putting in the variables into their Combination formula.

Also, thanks for the sarcastic response, appreciate it.

You post a thread which says how much you hate what you are doing, and post four questions without showing that you have made any effort (which is one of the rules of this forum) and you wonder why you get a sarcastic response?

But your idea of putting the variables into the combination formula is a good one, try it!

 

[MarkFL]

...Also, thanks for the sarcastic response, appreciate it.

Prove It did not post sarcastically...he was merely following our policy to sincerely ask users who posts questions without any work shown what they have tried. This helps us help you more efficiently, if we know where you are with the material. It is why we have MHB Rule #11 and we also ask in MHB Rule #8 that you post no more than two question in your initial post. When you post more than that, threads can become convoluted as several people might be trying to help with different questions simultaneously in the same thread.

For the first question, try applying the binomial theorem to the expansion of:

\(\displaystyle (1+1)^n\)

 

[Greg]

For #2, you've got \(\displaystyle \dfrac{(n-1)!}{r!(n-1-r)!}+\dfrac{(n-1)!}{(r-1)!(n-r)!}\)

which may be written as

\(\displaystyle (n-1)!\left(\dfrac{(r-1)!(n-r)!+r!(n-1-r)!}{r!(n-1-r)!(r-1)!(n-r)!}\right)\)

Now think of how, using the properties of factorials, you can rewrite the nominator as

\(\displaystyle (r-1)!\cdot(n-1-r)!\cdot n\).