Question

Counting Distributions. (a). How many solutions are there to the equation 1 + 12 +13 + 14 = 20 if ni, ng, 13, 14 are nonnegative integers and na 4? (b). How many ways are there to give 13 (identical) red pencils and 15 (identical) blue pencils to 11 (different) students? (c). What if each student needs at least 1 pencil of each color?

Answer 1

Solution:-

Given that

(a)

...........(1)

ni+n2 +ng+n4 20

Note: Number of non-negative integer solution of

...........(2)

21 +92 +93 + ... + I'm n

is

n + m - 1 n

here basically we have to arrange n 1's and m-1 +'s . And every such arrangements/permutations

of those n 1's and m-1 +'s corresponds to a particular solution of (2) and vice versa.

So, number of solution of (2) is the number of permutations of n 1's and m-1 +'s,

i.e.,  $\frac{(n+m-1)!}{n!(m-1)!}=\begin{pmatrix} n+m-1\\ n \end{pmatrix}$

Thus number of non negative integer solution of (1) is

$\begin{pmatrix} 20+4-1\\ 20 \end{pmatrix}=\begin{pmatrix} 23\\ 20 \end{pmatrix}$

Now, if $n_4=4$ , then  $n_1+n_2+n_3=16$ .......(3)

So, number of solution of (1) when $n_4=4$ is the number of solution of (3)

i.e.,  $\begin{pmatrix} 16+3-1\\ 16 \end{pmatrix}=\begin{pmatrix} 18\\ 16 \end{pmatrix}$

Thus required number of solution is

$\begin{pmatrix} 23\\ 20 \end{pmatrix}-\begin{pmatrix} 18\\ 16 \end{pmatrix}$

= 1618

b)

Let
C
and $y_i$ be the number of red and blue pencils given to i th student, for i = 1, 2, 3, ..., 11

So, $x_i,y_i$ are non-negative integers, and

$x_1+x_2+....+x_{11}=13$ ..(1) [Total Red pencils = 13]

$y_1+y_2+....+y_{11}=15$ ....(2) [Total Blue pencils = 15]

Thus, required number of ways to distribute is

total number of solution of (1) * total number of solution of (2)

$\begin{pmatrix} 13+11-1\\ 13 \end{pmatrix}\times \begin{pmatrix} 15+11-1\\ 15 \end{pmatrix}$

  $=\begin{pmatrix} 23\\ 13 \end{pmatrix}\cdot \begin{pmatrix} 25\\ 15 \end{pmatrix}$

c)

Note:- Number of positive integer solution of

....(1)  

21 +92 +93 + ... + I'm n

is $\begin{pmatrix} n-1\\ m-1 \end{pmatrix}$

$(x_1-1)+(x_2-1)+...+(x_m-1)=n-m$

$y_1+y_2+...+y_m=n-m$ ...(2)

where $y_i=x_i-1$

for i = 1, 2, ..., m

So, $y_i$ are non-negative integer

Thus, number of positive integer solution of (1) is

the number of non negative solution of (2)

i.e., $\begin{pmatrix} n-m+m-1\\ n-m \end{pmatrix}=\begin{pmatrix} n-1\\ n-m \end{pmatrix}=\begin{pmatrix} n-1\\ m-1 \end{pmatrix}$

Here we have

$x_1+x_2+...+x_{11}=13$ .....(3)

$y_1+y_2+...+y_{11}=15$ .....(4)

as described in (b)

Thus required total number of distributions is

total number of positive solution of (5) x total number of positive solution of (4)

= $\begin{pmatrix} 13-1\\ 11-1 \end{pmatrix}\times \begin{pmatrix} 15-1\\ 11-1 \end{pmatrix}$

$=\begin{pmatrix} 12\\ 10 \end{pmatrix}\cdot \begin{pmatrix} 14\\ 10 \end{pmatrix}$

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