Suppose we want to choose 5 letters, without replacement, from 9 distinct letters. (If necessary, consult a list of formulas.) (a) How many ways can this be done, if order of the choices is not relevant? .

Answer:

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.

Step-by-step explanation:

This was someone else’s work not mine sorry here’s creditssss :))

Answer: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.

Step-by-step explanation:This was someone else’s work not mine sorry here’s creditssss :))

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