Main post link -> http://blog.brilliant.org/2013/01/30/combinations/

Read the full article at the blog, feel free to ask questions, discuss the topic, or the test yourself questions below

**Test yourself**

How many ways can three different appetizers be chosen from a menu that has 10 choices?

In New York city, all the streets are arranged in a grid. A hospital is located 5 blocks east and 6 blocks north of an accident. How many ways are there to get there, if we only go 1 block north or 1 block east at each intersection?

An office with 30 people wants to have 4 teams of 5 players participate in a charity tournament. How many ways can the teams be made? Hint: Be very careful.

How many ordered integer solutions \((a, b, c, d)\) are there to \(a + b + c + d = 20\) subject to \( a \geq 1, b \geq 2, c \geq 3, d \geq 4 \)?

Winston must choose 4 courses for his final semester of school. He must take at least 1 science class and at least 1 arts class. If his school offers 4 science classes, 3 arts classes and 3 other classes, how many different choices for classes does he have?

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## Comments

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TopNewestProblem 2 is 11C5

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sorry i made my calculation of first wrong the correct answer for 1 is 120 as 720/6

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The fourth answer should be 13C3 which is 13x12x11/6 which is 286.

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For fifth what is the correct answer? i got 672 as my answer, \(4*3*8*7\)

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The third should be 30C5 x 25C5 x 20C5 x 15C5. Four teams will be formed, with 5 players each. 10 people will be left over.

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This is actually a very common mistake made (also in this week's problem set). Remember that the order of the teams doesn't matter.

According to your method, how many ways are there for an office of 3 people to pick 3 teams of 1 person each? There should clearly only be 1 possible way.

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Is the answer \[\binom{30}{20} \times \frac{(20!)}{(5!)^{4} \times (4!)}\]

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I understand. What is the correct way to do this question?

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the answer of 1 is 720

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Answer for Question 1:720 ways because after you pick the first one,there are nine choices left.After you pick the second one there are 8 choices left so \(10 \times 9 \times 8 \) =720

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The answer is 120 because those ways of choosing the three appetizers are counted many times (ABC, ACB, BAC, BCA, CAB, CBA are the same).

The answer is \({10 \choose 3}\) = \(\frac {10*9*8}{3*2*1}\) = \(120\)

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Thanks for pointing that out.It should be \( 10 \times 9 \times 8 \) = 720 / 6 = 120

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