3 years ago

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Mathematics

1 answer:

Basile [38]3 years ago

6 0

Answer:

is it multiple choise

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The answer should be C. 1/12.

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Read 2 more answers

Suppose you choose a team of two people from a group of n > 1 people, and your opponent does the same (your choices are allow

jonny [76]

Answer:

The number of possible choices of my team and the opponents team is

$\left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right i^{3}$

Step-by-step explanation:

selecting the first team from n people we have $\left(\begin{array}{ccc}n\\1\\\end{array}\right) = n$ possibility and choosing second team from the rest of n-1 people we have $\left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1$

As { A, B} = {B , A}

Therefore, the total possibility is $\frac{n(n-1)}{2}$

Since our choices are allowed to overlap, the second team is $\frac{n(n-1)}{2}$

Possibility of choosing both teams will be

$\frac{n(n-1)}{2} * \frac{n(n-1)}{2} \\\\= [\frac{n(n-1)}{2}] ^{2}$

We now have the formula

1³ + 2³ + ........... + n³ = $[\frac{n(n+1)}{2}] ^{2}$

1³ + 2³ + ............ + (n-1)³ = $[x^{2} \frac{n(n-1)}{2}] ^{2}$

= $\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] = [\frac{n(n-1)}{2}]^{3}$

4 0

4 years ago

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