3 years ago

Write two different ways to decompose 6/10 into a sum

Mathematics

1 answer:

Papessa [141]3 years ago

8 0

3/10 + 3/10

2/10 + 2/10 + 2/10

5/10 + 1/10

6/20 + 6/20

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Answer:

12.56 yards

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Step-by-step explanation:

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Compare 6,200.ft by 1.mi 900.ft

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Glad I could help, and good luck!

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15.

boyakko [2]

Explanation:

${3 \choose 2} = \frac{3!}{2!(3-2)!} = \frac{6}{2*1} = 3$

${3 \choose 1} = \frac{3!}{1!(3-1)!} = \frac{6}{1*2} = 3$

b) First, we have that

${n \choose k} = \frac{n!}{k!(n-k)!}$

On the other hand,

${n \choose n-k} = \frac{n!}{(n-k)!(n(n-k))!} = \frac{n!}{(n-k)!k!}$

Therefore, both expressions are equal. This makes sense, because selecting k elements from a group of n is the same than specify which elements you will not select. In order to specify those elements you need to select the n-k elements that will not be selected. Hence, each time you are selecting k from n, you are also selecting n-k from n, and from that reason both combinatorial numbers are equal.

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