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2 years ago

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Suppose we want to choose 6 letters, without replacement, from 15 distinct letters. (A) how many ways can this be done, if the o

rder of choices is not taken into consideration? (B) How many ways can this be done, if the order of choices is taken into consideration?

1 answer:

Olenka [21]2 years ago

7 0

Answer:

Below in bold.

Step-by-step explanation:

A.This is the number of combinations of 6 from 15

= 15C6

= 15! / (15-6)! 6!

= 5,005 ways.

B. This is the number of permutaions of 6 from 15:

= 15! / (15-6)!

= 3,603,600 ways.

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Answer:

Their would be 12 green skittles.

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3 years ago

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Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

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Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

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Replace y with a'(d)

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To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

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$a(a'(d)) = d$

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$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

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2 years ago

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Svetradugi [14.3K]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

because

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hope this helps

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2 years ago