Question

There are 5 white 4yellow, 3 green , 2blue and 1 red ball. The balls are identical except colour. These are to be arranged in a line at 5 places . Find the no. Of distinct arangement.

Answer 1

Answer on Question #56355 – Math – Combinatorics | Number Theory

There are 5 white, 4 yellow, 3 green, 2 blue and 1 red ball. The balls are identical except color. These are to be arranged in a line at 5 places. Find the no. Of distinct arrangement.

Solution

Cases are as follows:

1) 1 1 1 1, All balls are different colors

2) 2 1 1 1, Two balls are the same color, three balls are different colors

3) 2 2 1, Two balls are the same color, two balls are the same color, one ball is different colors

4) 2 3, Two balls are the same color, three balls are the same color

5) 3 1 1, Three balls are the same color, two balls are different colors

6) 4 1, Four balls are the same color, one ball is different colors

7) 5, All balls are the same color.

Corresponding numbers of arrangements:

1) 5 different balls can be placed in 5 positions in

$$5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5! = 120$$

ways

2) Pick the color of two balls from w, y, g, b: $\binom{4}{1}$,

pick other 3 different balls from the rest colors: $\binom{4}{3}$

place 5 balls in 5 positions: 5!

Two balls are the same so we should divide by 2!

Thus

$$\frac {\binom {4} {3} \binom {4} {1} 5 !}{2 !} = 960,$$

And so on

3) $\frac{\binom{4}{2}\binom{3}{1}5!}{2!2!} = 540,$

4) $\frac{\binom{3}{1}\binom{3}{1}5!}{2!3!} = 90,$

5) $\frac{\binom{3}{1}\binom{4}{2}5!}{3!} = 360,$

6) $\frac{\binom{2}{1}\binom{4}{1}5!}{4!} = 40,$

7) 1.

So $N = 120 + 960 + 540 + 90 + 360 + 40 + 1 = 2111$$.

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