Mathematics: Suppose we want to choose 6 letters, without replacement, from 15 distinct letters. (A) how many ways can this be done, if the order of choices is not taken into consideration? (B) How many ways can this be done, if the order of choices is taken into consideration?

Suppose we want to choose 6 letters, without replacement,

Ответ:

carlo123

21.01.2022

Mathematics

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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Ответ:

paulasuarez8276

03.08.2022

Mathematics

12 x + 1 = 3 ( 4 x + 1 ) - 2

12 x + 1 = 3*4 x + 3 * 1 - 2

12 x + 1 = 12 x + 3 - 2

12 x - 12 x = 3 - 2 - 1

0 x = 1 - 1

0 x = 0

True for all x

hope this helps!

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

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