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Suppose that 18 red beads, 12 yellow beads, eight blue beads, and 12 black beads are to be strung in a row. How many different arrangements of the colors can be formed?

Short Answer

Expert verified

The colors can be formed in \(5.134987617 \times {10^{26}}\) no. of arrangements.

Step by step solution

01

Given information

Given that, total no. of red beads \(\left( {{n_1}} \right)\)=18

Total no. of yellow beads \(\left( {{n_2}} \right)\)=12

Total no. of blue beads \(\left( {{n_3}} \right)\)=8

Total no. of black beads \(\left( {{n_4}} \right)\)=12

Total no. of beads

\(\begin{aligned}{}n &= {n_1} + {n_2} + {n_3} + {n_4}\\ &= 18 + 12 + 8 + 12\\ &= 50\end{aligned}\)

02

Compute the possible assignment

The Multinomial coefficient is

\(\left( {\begin{array}{*{20}{c}}n\\{{n_1},{n_2},{n_3}}\end{array}} \right) = \frac{{n!}}{{{n_1}!{n_2}!{n_3}!}}\)

Here,

\(\begin{aligned}{}\left( {\begin{array}{*{20}{c}}{50}\\{18,12,8,12}\end{array}} \right) &= \frac{{50!}}{{18!12!8!12!}}\\ &= 5.134987617 \times {10^{26}}\end{aligned}\)

Therefore, the colors can be formed in \(5.134987617 \times {10^{26}}\) no. of arrangements.

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Most popular questions from this chapter

Let \({A_1},{A_2}, \ldots \) be an arbitrary infinite sequence of events, and let \({B_1},{B_2}, \ldots \)be another infinite sequence of events defined as follows: \({B_1} = {A_1}\), \({B_2} = {A_1}^c \cap {A_2}\), \({B_3} = {A_1}^c \cap {A_2}^c \cap {A_3}\), \({B_4} = {A_1}^c \cap {A_2}^c \cap {A_3}^c \cap {A_4}\),…Prove that

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and that

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