Answer to Question #282374 in Discrete Mathematics for Neha

Question

Use the binomial theorem to find the coefficient of x^a y^b in the expansion of (2x^3

− 4y^2)

7, where

a) a = 9, b = 8.

b) a = 8, b = 9.

c) a = 0, b = 14.

d) a = 12, b = 6.

e) a = 18, b = 2.

35) How many unique partitions of the word ARKANSAS are there?

36) A group of six students consists of 3 seniors, 2 juniors, and 1 sophomore. How

many unique partitions of this group of students are there by grade?

Answer 1

34)

binomial theorem:

$(p+q)^n=\sum \begin{pmatrix} n \\ k \end{pmatrix}p^{n-k}q^k$

we have:

$(2x^3-4y^2)^7$

a)

coefficient of $x^9 y^8$ :

$\begin{pmatrix} 7 \\ 4 \end{pmatrix}(2x^3)^3(4y^2)^4=35\cdot8\cdot256x^9y^8=71680x^9y^8$

b)

coefficient of $x^8 y^9$ is 0, because it is impossible to get x⁸ having initial term x³ (same for y)

c)

coefficient of $x^0 y^{14}$ :

$\begin{pmatrix} 7 \\ 7 \end{pmatrix}(2x^3)^0(4y^2)^7=4^7y^{14}=16384y^{14}$

d)

coefficient of $x^{12} y^{6}$ :

$\begin{pmatrix} 7 \\ 3 \end{pmatrix}(2x^3)^4(4y^2)^3=35\cdot16\cdot64x^{12}y^6=35840x^{12}y^6$

e)

coefficient of $x^{18} y^{2}$ :

$\begin{pmatrix} 7 \\ 1 \end{pmatrix}(2x^3)^6(4y^2)^1=7\cdot64\cdot4x^{18}y^2=1792x^{18}y^2$

35)

a partition of a set is a grouping of its elements into non-empty subsets, in such a way that every element is included in exactly one subset.

A multinomial coefficient describes the number of possible partitions of n objects into k groups of size n₁, n₂, …, n_k.

Multinomial Coefficient = $\frac{n!}{n_1!n_2!...n_k!}$

for word ARKANSAS:

Multinomial Coefficient = $\frac{8!}{3!1!1!1!2!}=3360$

There are 3,360 unique partitions of the word ARKANSAS.

36)

Multinomial Coefficient = $\frac{6!}{3!2!1!}=60$

There are 60 unique partitions of these students by grade.