Permutation and Combination problem

kd15

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Dec 13, 2020
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Solve for n: 6C4 = 3 * (nC1) + (n-1)C(n-2)
Screenshot 2020-12-13 at 3.43.20 PM.pngScreenshot 2020-12-13 at 3.45.19 PM.pngScreenshot 2020-12-13 at 3.55.33 PM.png

When I say first image I will be referring to the picture with the green. When I say second image I am referring to the one with blue, red, and pink in it. The third image is the one that is mostly comprised of the yellow writing.

In the first image I solved for the left side, and got 15. I am about 99% sure that is correct. Then, on the right side (still talking about the first image), I simplified. I simplified to what I have on the second image. I am stuck here though. I am unsure of how you are supposed to solve for n. I rewrote the original problem but used the new n,r I had. That's what the bottom part of my image two is. I can't get passed that point, but I need to solve for n. I was able to solve, just through a guess and check method, but is there a way to do it not by just guessing? I got the answer of 4. The third image is showing my work when I put 4 in for n, and showing how it is correct because it is equal to 15, which is what I found 6C4 to be earlier. I just am unsure how to get to the answer through using math, and not simply just guessing.
 
[MATH]\binom{6}{4} = 3 \cdot \binom{n}{1} + \binom{n-1}{n-2}[/MATH]
[MATH]\dfrac{6!}{4!(6-4)!} = 3 \cdot \dfrac{n!}{1!(n-1)!} + \dfrac{(n-1)!}{(n-2)! [(n-1)-(n-2)]!}[/MATH]
[MATH]15 =3n + (n-1)[/MATH]
[MATH]16 = 4n \implies n = 4[/MATH]
 
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