Question

Given 5 different green dyes, four different blue dyes and three different red dyes, how many combinations of dyes can be chosen taking at least one green and one blue dye?

Answer 1

Hint- Here, we will proceed by evaluating the number of combinations of choosing at least one green dye, at least one blue dye, all the combinations possible for red dyes separately using the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] and then multiplying these values.

Complete step-by-step solution -

Given, Number of different green coloured dyes = 5
Number of different blue coloured dyes = 4
Number of different red coloured dyes = 3
Therefore, total number of different coloured dyes = 5+4+3=12
Ways of selecting at least 1 green dye out of 5 different green dyes includes all the cases of selecting more than 1 green dye (i.e., 1 or 2 or 3 or 4 or 5).
As we know that the number of ways of selecting r items out of n items is given by \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Number of ways of selecting at least 1 green dye out of 5 different green dyes \[
   = {}^5{C_1} + {}^5{C_2} + {}^5{C_3} + {}^5{C_4} + {}^5{C_5} \\
   = \dfrac{{5!}}{{1!\left( {5 - 1} \right)!}} + \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}} + \dfrac{{5!}}{{3!\left( {5 - 3} \right)!}} + \dfrac{{5!}}{{4!\left( {5 - 4} \right)!}} + \dfrac{{5!}}{{5!\left( {5 - 5} \right)!}} \\
   = \dfrac{{5.4!}}{{1!4!}} + \dfrac{{5.4.3!}}{{2!3!}} + \dfrac{{5.4.3!}}{{3!2!}} + \dfrac{{5.4!}}{{4!1!}} + \dfrac{{5!}}{{5!0!}} \\
   = 5 + 10 + 10 + 5 + 1 \\
   = 31 \\
 \]
\[ \Rightarrow \]Number of ways of selecting at least 1 green dye out of 5 different green dyes = 31
Now, ways of selecting at least 1 blue dye out of 4 different blue dyes includes all the cases of selecting more than 1 blue dye (i.e., 1 or 2 or 3 or 4).
Number of ways of selecting at least 1 blue dye out of 4 different blue dyes \[
   = {}^4{C_1} + {}^4{C_2} + {}^4{C_3} + {}^4{C_4} \\
   = \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} + \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} + \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} + \dfrac{{4!}}{{4!\left( {4 - 4} \right)!}} \\
   = \dfrac{{4.3!}}{{1!3!}} + \dfrac{{4.3.2!}}{{2!2!}} + \dfrac{{4.3!}}{{3!1!}} + \dfrac{{4!}}{{4!0!}} \\
   = 4 + 6 + 4 + 1 \\
   = 15 \\
 \]
\[ \Rightarrow \]Number of ways of selecting at least 1 blue dye out of 4 different blue dyes = 15
Now, the total number of ways possible for selection from 3 different red dyes includes all the cases possible for selection of red dyes (i.e., 0 or 1 or 2 or 3).
Number of ways of selecting red dye/dyes out of 3 different red dyes \[
   = {}^3{C_0} + {}^3{C_1} + {}^3{C_2} + {}^3{C_3} \\
   = \dfrac{{3!}}{{0!\left( {3 - 0} \right)!}} + \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} + \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}} + \dfrac{{3!}}{{3!\left( {3 - 3} \right)!}} \\
   = \dfrac{{3!}}{{0!3!}} + \dfrac{{3.2!}}{{1!2!}} + \dfrac{{3.2!}}{{2!1!}} + \dfrac{{3!}}{{3!0!}} \\
   = 1 + 3 + 3 + 1 \\
   = 8 \\
 \]
\[ \Rightarrow \]Number of ways of selecting red dye/dyes out of 3 different red dyes = 8

Thus, total number of combinations of dyes chosen taking at least one green and one blue dye is given by
Total number of required combinations $ = 31 \times 15 \times 8 = 3720$ ways.

Note- In this problem, the number of ways of selecting at least 1 green dye out of 5 different green dyes can be obtained by ${2^5} - {}^5{C_0}$ and the number of ways of selecting at least 1 blue dye out of 4 different blue dyes can be obtained by \[{2^4} - {}^4{C_0}\] and the total number of ways of selection of red dyes out of 3 different red dyes can be obtained by \[{2^3}\].