Question

In how many ways can $5$ colors be selected out of $8$ different colors including red, blue and green $(a)$ if blue and green are always to be included $(b)$ if red is always excluded. $(c)$ If red and blue are always included but green excluded?

Answer 1

Hint: To solve this type of question we should know the details of the non-fixed objects or places of the colors and the colors which are always excluded or included.
So here we are needed to select the colors by using the given data.
Then we have to find one by one by using a combination formula.
Finally we get the required answer.

Formula used: The number of ways of selecting the colors by using the formula ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ we will have answered.

Complete step-by-step answer:
In this question it is given that there are $5$ colors to be selected out of $8$ different colors including red, blue and green.
In case, when we are selecting blue and green colors already.
$5$ Colors $ \to $$2$ Colors are already selected (given in the question)
Remaining colors are to be chosen is $3$
Therefore, the required selection is ${}^6{C_3}$
Here \[n = 6\] and \[r = 3\]
We use the formula and we write it as,
$\Rightarrow$${}^6{C_3} = \dfrac{{6!}}{{3!(6 - 3)!}}$
On subtracting the bracket term we get,
$\Rightarrow$${}^6{C_3} = \dfrac{{6!}}{{3!(3)!}}$
On splitting the factorial we get
$\Rightarrow$${}^6{C_3} = \dfrac{{6 \times 5 \times 4 \times 3!}}{{3!\left( 3 \right)!}}$
On cancelling the term and also split the factorial we get,
$\Rightarrow$${}^6{C_3} = \dfrac{{6 \times 5 \times 4}}{{3 \times 2 \times 1}}$
Let us divide the term and we get
$\Rightarrow$${}^6{C_3} = 2 \times 5 \times 2$
On multiplying the terms we get
$ = 20$
In case of red color is excluded
$7$ Colors are remaining
Hence, the required selection is
$ \Rightarrow {}^7{C_5}$
Here \[n = 7\] and \[r = 5\]
We use the formula and we write it as,
$\Rightarrow$${}^6{C_3} = \dfrac{{7!}}{{5!(7 - 5)!}}$
On subtracting the bracket term we get,
$\Rightarrow$${}^6{C_3} = \dfrac{{7!}}{{5!(2)!}}$
On splitting the factorial we get
$\Rightarrow$${}^6{C_3} = \dfrac{{7 \times 6 \times 5!}}{{5!\left( 2 \right)!}}$
On cancelling the term and also split the factorial we get,
$\Rightarrow$${}^6{C_3} = \dfrac{{7 \times 6}}{{2 \times 1}}$
Let us divide the term and we get
$\Rightarrow$${}^6{C_3} = 7 \times 3$
Let us multiply the terms we get,
$ = 21$
We have total $8$ colors red and blue are always included, so $2$ out of $5$ spaces are filled.
We need to fill $3$ more spaces we have $6$ colors left but green color is always excluded, so we have to choose from only $5$ colors to fill $3$ places which is ${}^5{C_3}$
Here \[n = 5\] and \[r = 3\]
We use the formula and we write it as,
$\Rightarrow$${}^6{C_3} = \dfrac{{5!}}{{3!(5 - 3)!}}$
On subtracting the bracket term we get,
$\Rightarrow$${}^6{C_3} = \dfrac{{5!}}{{3!(2)!}}$
On splitting the factorial we get
$\Rightarrow$${}^6{C_3} = \dfrac{{5 \times 4 \times 3!}}{{3!\left( 2 \right)!}}$
On cancelling the term and also split the factorial we get,
$\Rightarrow$${}^6{C_3} = \dfrac{{5 \times 4}}{{2 \times 1}}$
Let us divide the term and we get
$\Rightarrow$${}^6{C_3} = 5 \times 2$
Let us multiply the terms we get,
$ = 10$

Hence required selection is ${}^5{C_3}$$ = 10$

Note: In the total number of ways of dividing \[\;n\] identical things among\[\;r\] persons such that each one gets at least one is ${}^{n - 1}{C_{r - 1}}$
The total number of combinations of \[\;n\] different objects taken \[\;r\] at a time in which as follows:
\[\left( 1 \right)\;\] If we take\[\;m\] particle objects are excluded then we can write it as, ${}^{n - m}{C_r}$
\[\left( 2 \right)\] If we take\[\;m\] particular objects are included then we can write it as, ${}^{n - m}{C_{r - 1}}$.