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| Quiz of the Month (January 2007)
Hector C. Parr
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SOLUTION TO LAST MONTH'S QUIZ
1. 15
2. 511
3. 9330
Notes.
Q.2 Temporarily label the groups X and Y.
Each person goes into group X or Y.
So total number of arrangements = 210 = 1024
But this includes cases where X or Y is empty.
So number of arrangements = 1024-2 = 1022 - - A
But on removing group labels, groups may be interchanged.
So number of different arrangements = 1022/2 = 511
Q.3 Temporarily label the groups X, Y and Z.
Each person goes into group X or Y or Z.
So total number of arrangements = 310 = 59,049
But this includes cases where one group is empty, and all
people are distributed among other two groups
i.e. 3*1022 = 3066 cases (see A above)
It also includes 3 cases where two groups are empty.
So total number of arrangements = 59,049-3066-3 = = 55,980
On removing group labels groups may be rearranged in 6 ways.
So number of different arrangements = 55,980/6 = 9330
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THIS MONTH'S QUIZ
A street consists of 20 houses, and the numbers of children
living in each house are as follows:
No. of children in house: 0 1 2 3 4
No. of houses: 4 6 5 3 2
1. If a house is chosen at random, what is the probability
that exactly 3 children live there?
2. If a child is chosen at random, what is the probability that
he or she lives in a house where exactly 3 children live?
3. If two different houses are chosen at random, what is the
probability that they are homes to the same number of children?
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© Hector C. Parr (2007)
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