# Multinomial Theorem - Hemant Malhotra

• (x+a)^n = nC0 * x^n + nC1 * x^(n-1) * a + nC2 *(x)^(n-2) * a^2 ... nCn * a^n

note- total number of terms here n+1 not n . bcz n is varying from o to n so total terms will be n+1

So general term (rth term of this expansion) of this expansion is T(r+1) = nCr * x^r * (a)^(n-r) where r varies from 0 to n

Consider series (x + y + z + k)^n
So general term of this equation will be = [ n!/(r1! * r2! * r3! * r4!)] * (x)^r1* (y)^r2 * (z)^r3 * (k)^r4
where r1 + r2 + r3 + r4 = n

Find coefficient of x^3 * y^4 * z^3 in expansion of (2x + y + 4z)^10

general term of this equation is = 10!/(r1! * r2! * r3!) * (2x)^r1 * (y)^r2 * (4z)^r3
= 10!/(r1! * r2! * r3!) * (2)^r1 * (4)^r3 * x^r1 * y^r2 * z^r3
now we have to find x^3 * y^4 * z^3 so r1=3 ,r2=4 and r3=3
put tvalue of r1, r2, r3 here 10!/(r1! * r2! * r3!) * (2)^r1 * (4)^r3 * x^r1 * y^r2 * z^r3
so coefficient is 10!/(3! * 4! * 3!) * (2)^3 * (4)^3

Find the greatest coefficient in (x + y + z + k)^15

general term = 15!/(r1! * r2! * r3! * r4!) * x^r1 * y^r2 * z^r3 * k^r4
so basically we have to maximize 15!/(r1! * r2! * r3! * r4!) term
so we have to minimize (r1! * r2! * r3! * r4!) so for this symmetry is useful so put values as close as possible here
r1 = r2 = r3 = 4 and r4 = 3
so greatest coefficient is 15!/(3! * (4!)^3)

find coefficient of x^6 in (1 + x^2 - x^3)^8

general term = 8!/r1! * r2! * r3! * (1)^r1 * (x^2)^r2 * (-x^3)^r3
= 8!/r1! * r2! * r3! * x^(2r1) * (-1)^r3 * (x)^3r3
= 8!/r1! * r2! * r3! * (-1)^r3 * x^(2r2+3r3)
now we have to find coefficient of x^6 here (for any coefficient same procedure )
so r1 + r2 + r3 = 8
2r2 + 3r3 = 6
find r1, r2 and r3 here
let r3=0 so r2=3 so r1=2 and let r3=2,so r2=0 so r1=6 (r1,r2,r3 should be integer because term can't be negative)
now coefficient will be 8!/r1! * r2! * r3! *(-1)^r3
put both values of r1,r2 and r3 and add
so 8!/5! * 3!+8!/2! * 6!=84
NOTE- method seems lengthy but it will hardly take 1 to 2 min with 100% accuracy (y)

CONCEPT-

coefficient of x^r in (1-x)^-n is n+r-1cn-1 (note that it looks like n+r-1cr-1 but its n+r-1cn-1 in this case)

Find coefficient of x^13 in (x+x^2+x^3+x^4+x^5+x^6)^4

(x + x^2 + x^3 + x^4 + x^5 + x^6)^4
= x^4(1 + x + x^2 + x^3 + x^4 + x^5)^4
=x^4(1-x^6/(1-x) )^4
= x^4(1-x^6)^4 (1-x)^-4
so coefficient of x^13 in x^4(1-x^6)^4 (1-x)^-4
so coefficient of x^9 in (1-x^6)^4 (1-x)^-4
(1-x^6)^4 general term of this is 4cr * (1)^r * (-x^6)^(4-r)= 4cr * (-1)^(4-r) * x^(24-6r)
(1-x)^-4 coefficient of x^r1 is 4+r-1c4-1=(3+r1)C3
so 4cr * (-1)^(4-r)*x^24-6r+r1 *( 3+r1)C3 so for coefficient of x^9
24-6r+r1=9 so 6r-r1=15 so r =3 and r1=3 or r=4 and r1=9
so coefficient = 4cr * (-1)^(4-r) * ( 3+r1)C3 ..
= 4c3 * (-1)^1 * 6c3 + 4c4 * (-1)^0 * (3+9)c3
= 140

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