Stuck on this combinatorics question

Question

A florist sells roses of five different colors. How many bunches of a half-dozen roses can be formed?

 

a 196

2. 210
3. 236
4. 300

Please help me understand this question with proper explanation. I got the answer but looking for the logical reasoning to it.

Comments

Let's understand the question.

Here there are five different types of colors. we have to form the bunches of half-dozen. 

So, here the number of roses are infinite. this question is nothing but the finding of the non-negative solution of 

                                X1 + X2 + X3 + X4 + X5 = 6

$\binom{6+5-1}{5-1}$

$\binom{10}{4}$ = 210.

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Please explain this 6 + 5 -1 and 5 - 1. How you got this?

Answer 1

Let 5 colors be $\color{red}{ \text{Red}}, \color{blue}{ \text{Blue}},  \color{green}{ \text{Green}},  \color{orange}{ \text{Orange}},  \color{Black}{ \text{Black}}  $

Florist have as many roses as your want of these 5 colors

We need to form bunch of half dozen (6)

$\color{red}{ \text{Red}}+ \color{blue}{ \text{Blue}}+  \color{green}{ \text{Green}}+  \color{orange}{ \text{Orange}}+  \color{Black}{ \text{Black}} = 6  $

using stars and bars,

$\large \binom{6+5-1}{5-1} = \binom{10}{4} = 210$

Comments

@Lakshman Patel RJIT Definitely bro.

I forget to select yours as well. My mistake will do it now. 

Thanks

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The answer is correct for sure .But here it should be ( 10 C 6 ) rather than ( 10 C 4). it is just coincidence that ans matches (as 4+6=10) .

basically this is a problem of selecting 'r' objects among 'n' different objects with replacement.

r=6 and n=5

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see the above link you will definitely understand.

$\binom{n}{r}=\binom{n}{n-r}$

i.e. $n=1,r=0$

 $\binom{1}{0}=\binom{1}{1-0}$     

 $\binom{1}{0}=\binom{1}{1}$ 

Using star and bars

$x_{1}+x_{2}+x_{3}+.......+x_{n}=k$  where $x_{1},x_{2},x_{3},....x_{n}\geq0$

Number of integral solutions$=\binom{n+k-1}{k}=\binom{n+k-1}{(n+k-1)-k}=\binom{n+k-1}{n-1}$