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fgiga [73]
3 years ago
5

How many ways are there to choose a committee of 7 people from a group of 10 people?7201208405040?

Mathematics
1 answer:
lyudmila [28]3 years ago
5 0

Answer: 120

---------------------------------------------------------------------------------------

Explanation:

We have seven slots to fill. Let's call the slots A, B, C, D, E, F, G

There are 10 choices for slot A

There are 9 choices for slot B

There are 8 choices for slot C

There are 7 choices for slot D

There are 6 choices for slot E

There are 5 choices for slot F

There are 4 choices for slot G

We count down because each previous slot reduces the pool to choose from. Multiply those values out: 10*9*8*7*6*5*4 = 604800

Now divide that result by 5040. Why 5040? Because this is the number of ways to arrange 7 people within a single grouping. Note how 7! = 7*6*5*4*3*2*1 = 5040. W're dividing because order does not matter for any committee.

So we have 604800/5040 = 120 which is the answer we want

---------------

Side note: We could use the alternate route of the nCr combination formula

nCr = (n!)/(r!*(n-r)!)

where n = 10 and r = 7


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Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

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\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

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\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

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\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

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We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

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at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

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14=\dfrac{8x}{27} - \dfrac{38}{27}

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HI Sydney!

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