the easy way to expand (a+b+c)^n, the trinomial theorem

Here's the easy way to expand (a+b+c)^n, the trinomial theorem. You would see this in a discrete math class.

Binomial and trinomial theorem:

#combinatorics #discretemath

[tarunpurohit]

The greatest thing about you is u teach maths in a gentle way that is not painfull you do hard problem with such an ease and with having smile on your face.
Black pen red pen yeah😊

[noobkilla]

oh man, you just made discrete math soooo much more easier and fun with your previous 3 videos. How knew it wasn't so complicated after all?
I can't thank you enough!

[shivimish]

Can you also make a video on the multinomial theorem? 3 IS NOT ENOUGH !😂

[WerewolfLord]

I saw i j k, was getting excited for quaternions.

[David-wwsg]

I got 36 terms... (Correct me if I'm wrong!)

(a+b+c)^7 = [a+(b+c)]^7, so just treat (b+c) as a single value to make things easier.

Expand this out using the binomial theorem (I'm gonna ignore the co-efficients because they don't really matter):

a^7 + a^6 (b+c) + a^5 (b+c)^2 + a^4 (b+c)^3 + a^3 (b+c)^4 + a^2 (b+c)^5 + a (b+c)^6 + (b+c)^7

By observation we know the number of terms from (p + q)^n is gonna be n+1 terms.
So, looking at each term here, we know there's gonna be 1 term in a^7, 2 terms in a^6 (b+c), 3 terms in a^5 (b+c)^2, etc.
Thankfully, there will be no 'merging' or simplifying of the terms because they're multiplied by different powers of a, so adding these together gives:
1+2+3+4+5+6+7+8 = 36 terms.

[MrJdcirbo]

Holy moly. You just clarified combinatorics for me. The actual formula is n choose a = n!/(a!(n-a)!). It never occurred to me that this simplified to a terms of n! from n down. Only a numbers are multiplied. That's a big help. Thank you!

[vighnesh]

You just proved the question I had for 5 years: Proof of binomial and multinomial theorem in the starting 3.5 minutes. I love you for that.

[neilgerace]

8:34 "don't get too excited" :)

[guilhermeneryrocha]

Thank you so much for these videos. I love your channel but sometimes I think that all you do is integrals and I love integrals but some change on the subject is always good. Keep up this good work, man

[deepakishere]

Sir
you teach us Permutations from Combinations.
Thank tou sir🙏🙏

[nicolasgoubin]

Oh my ! The lengendary one... the PURPLE PEN :o

[tanushjain]

Thank you so much blackpenredpen my teacher gives these annoying trinomial expansions which took a lot of time to solve using only binomial theorem but now i can quickly solve these questions

[DavidS-qnjm]

I did it by having (a+b+c)^7 = (a+y)^7 where y=(b+c) to get the coefficient of a^3 y^4 as 35, then finding the coefficient of b^2 c^2 in y^4 to ultimately get 210 too. I imagine this method breaks down somewhat if you had a variable appearing twice inside the bracket such as (a+ab+c)^7 though.

For how many terms, I started with a^7, one term. For a^6, you can have b^1c^0 and b^0c^1, 2 more terms. For a^5 you have b and c ranging from 0 to 2, so 3 more terms. And so on until you have 8 different terms with a^0, with b and c ranging from 0 to 7. Thus 1+2+3+4+5+6+7+8 = 36.

[cicik]

this explanation with places is realy awsome, now it is simple to generalize that to any amount of summants

[msolec]

Stars and Bars theorem says 9c2 terms (36).

[marpin]

A general formula for the number of terms in (x+y+z)^n is (n+1)(n+2)/2

[kamarinelson]

I grouped it as (a+(b+c))^7. From there I know nCr(7, 3)*a^(3)*(b+c)^(4) gives me all the a cubed terms. Because I specifically only want the coefficient of the a^(3)*b^(2)*c^(2) term, I know I need nCr(7, 3)*a^(3)*nCr(4, 2)*b^(2)*(c)^2 = 210.

[samihenrisubi]

Also n+2 choose 2 gives the number of terms for the n power of the trinomial and for n= 7
9 choose 2

[patricemudimu]

Thank you for the video it has helped me a lot

[amankashyap]

i did in my head using multinomial theorem, but watching this also fun.