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In how many ways can 2 red, 2 black, 3 white and 2 blue balls be selected from 4 red, 3 black, 4 white and 8 blue balls? In how many ways can they be arranged?

Respuesta :

LammettHash
From the total pool of colored balls, one can choose 2 reds, 2 blacks, 3 whites, and 2 blues in

[tex]\dbinom42\cdot\dbinom32\cdot\dbinom43\cdot\dbinom82=6\cdot3\cdot4\cdot28=2016[/tex]

ways.

I'm assuming no ball of the same color is distinguishable from any other ball of the same color. So when I'm considering the possible arrangements, if I had lined up the ball as

red1 - black - red2 - ...

then this would be no different that

red2 - black - red1 - ...

So I now have 9 balls to arrange, which means there are [tex]9!=362,880[/tex] total possible permutations of them. But order among distinct colors is assumed to not matter. This means I have to divide the total number of permutations by the number of ways I could permute balls of the same color. Then there would be a total of

[tex]\dfrac{9!}{2!\cdot2!\cdot3!\cdot2!}=7,560[/tex]

ways of arranging the balls I had selected.
MrRoyal

The ways of arrangement of the balls are illustrations of combinations.

The number of selection is 2016

The given parameters are:

Total number of balls

[tex]\mathbf{Blue = 8}[/tex]

[tex]\mathbf{White = 4}[/tex]

[tex]\mathbf{Black = 3}[/tex]

[tex]\mathbf{Red = 4}[/tex]

Selected balls

[tex]\mathbf{Blue = 2}[/tex]

[tex]\mathbf{White = 3}[/tex]

[tex]\mathbf{Black = 2}[/tex]

[tex]\mathbf{Red = 2}[/tex]

So, the number of selection is:

[tex]\mathbf{n =^8C_2 \times ^4C_3 \times ^3C_2 \times ^4C_2}[/tex]

Apply combination formula, we have:

[tex]\mathbf{n =28 \times 4 \times3 \times 6}[/tex]

[tex]\mathbf{n =2016}[/tex]

Hence, the number of selection is 2016

Read more about combinations at:

https://brainly.com/question/4546043

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