There are 6 balls, all uniquely coloured, in a bag. We're randomly drawing 2. How many ways can we draw first Blue, then White? How many ways can we draw the colour combination of Blue and White?

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There are 6 balls, all uniquely coloured, in a bag. We're randomly drawing 2. How many ways can we draw first Blue, then White? How many ways can we draw the colour combination of Blue and White?

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Answer 1 · 2017-03-14T02:00:28

$P_{6, 2} = 30; C_{6, 2} = 15$ $P_(6,2)=30; C_(6,2)=15$

Explanation:

We have 6 marbles and are pulling out 2.

The first question asks about colour sequences (and so Blue, White is different than White, Blue) and so is a permutation. The general formula is:

$P_{n, k} = \frac{n!}{(n - k)!}; n = population, k = picks$ $P_(n,k)=(n!)/((n-k)!); n="population", k="picks"$

$P_{6, 2} = \frac{6!}{(6 - 2)!} = \frac{6!}{4!} = \frac{6 \times 5 \times 4!}{4!} = 30$ $P_(6,2)=(6!)/((6-2)!)=(6!)/(4!)=(6xx5xx4!)/(4!)=30$

The second part of the question asks about colour combinations, and so Blue, White is the same as White, Blue. That is a combination and the general formula is:

$C_{n, k} = \frac{n!}{(k!) (n - k)!}$ $C_(n,k)=(n!)/((k!)(n-k)!)$ with $n = population, k = picks$ $n="population", k="picks"$

$C_{6, 2} = \frac{6!}{(2!) (6 - 2)!} = \frac{6!}{(2!) (4!)} = \frac{6 \times 5 \times 4!}{2 \times 4!} = 15$ $C_(6,2)=(6!)/((2!)(6-2)!)=(6!)/((2!)(4!))=(6xx5xx4!)/(2xx4!)=15$

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There are 6 balls, all uniquely coloured, in a bag. We're randomly drawing 2. How many ways can we draw first Blue, then White? How many ways can we draw the colour combination of Blue and White?

1 Answer

Explanation:

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