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Andrei [34K]
3 years ago
11

seven different gifts are to be distributed among 10 children. How many distinct results are possible if no child is to receive

more than one gifts?
Mathematics
1 answer:
Alik [6]3 years ago
6 0

Answer:

120 distinct results are possible if no child is to receive more than one gifts.

Step-by-step explanation:

When the order is not important, we use the combination formula:

C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

The order is said to be not important if for example, John receiving the Buffalo Bills jersey and then Laura receiving the Cleveland Browns jersey is the same as Laura receiving the Cleveland Browns jersey before John receives the Buffalo Bills jersey.

In this problem, we have that:

Combinations of 7 from a set of 10 elements. So

C_{n,x} = \frac{n!}{x!(n-x)!}

C_{10,7} = \frac{10!}{7!(3)!} = 120

120 distinct results are possible if no child is to receive more than one gifts.

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Answer:

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Step-by-step explanation:

When working with surds we need to take note of the roots present there.

To expand this equation we can do it the following way noting that √3 X √3 = 3

<em></em>

<em>Expanding (1-√3)(⅓+√3)</em>

1 X 1/3 = 1/3

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hence, expanding the equation, we have

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Is means equals, of means multiply and percents need to be changed to decimals.

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