Question

In: Statistics and Probability

How many ways are there to choose 5 out of 10 homebuyers and 3 out of...

How many ways are there to choose 5 out of 10 homebuyers and 3 out of 8 realtors and have them stand in a line so that realtors do not stand next to each other?

Solutions

Expert Solution

Out of 10 homebuyers we have to choose 5 homebuyers. this is a question of selection that can be done in 10C5 ways.

Out of 8 realtors 3 has to be chosen again this a question of selection that can be done in 8C3 ways.

Now as we have to choose 5 homebuyers and 3 realtors ( as "and" is there intermidiate sign is multiplication) that can be done in 10C5 *  8C3 =14112   ways.

Now they have to be arranged in a straight line so that no realtors stand next to each other

R H R H R H R H R H R. This can be the diagram where they are standing in a straight line and no realtors are standing next to each other.  H is for home buyers and R is for realtors. As this is qusetion of arrangement we have to use permutation

There is no restriction for home buyers. They can be arranged in 5P5ways

There is restriction for realtors. From the above diagram it is clear that there are six places for them and they have to occupy three places. That can be done in 6P3 ways.

So thae total number of ways they can be arranged is 5P5  * 6P3 = 14400 ways

So the total possible ways is 14112 * 14400=203212800 ways


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