Answer to Question #188446 in Combinatorics | Number Theory for Geetika

Question

Consider the simple problem of placing four coloured balls: red, blue, green and white in 15

boxes. What are the numbers of distinct ways in which the balls can be placed in these

boxes, if each box can hold only one ball? Also write the generalized formula of this

numerical result.

Answer 1

we can solve it as follows using :

$\bigstar$if balls =4

answer 16 bacause a ball of each color may or may not be = 2 ^ 4

Generalized formula

$(1+n_r)*(1+n_b)*(1+n_g)*(1+n_w),if (n_r+n_b+n_g+n_w )\leq n_{box}$

$\bigstar$ if we have 4 type balls and each number =$n_{box}$ 

i.e. we have 15 red balls,we have 15 blue balls ,we have 15 geen balls and we have 15 white balls.

if we need not empty box

we have $C^{3}_{15}=816$ 

 because string a string of 18 characters (15 'b', 3 '|')bijection task

'b' before first '|' - red balls, between first and second '|'-blue balls, between second and thrid '|' -green balls ,after thrid '|' - white balls.

Generalized formula:

$C^{type\ color-1}_{n_{box}+type\ color-1}$