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?

Respuesta :

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.

Hence (A)ways can this be done, if the order of choices is not taken into consideration is 5005

(B) ways can this be done, if the order of choices is taken into consideration is 3,603,600

What is Permutation ?

A permutation is a mathematical calculation that determines the number of different ways a given set can be organized, with the order of the arrangements being important.

What is Combination?

A combination is a mathematical technique for calculating the number of potential arrangements in a set of things where the order of the items is irrelevant. You can choose the components in any order in combinations. Permutations and combinations are often mistaken.

How to solve?

No of letters to be chosen =6

Total no of letters=15

Formula used [tex]\frac{No of favourable outcomes }{Total no. of observations}[/tex]

=6/15

A)

If the order is not considered one can use combination in order to obtain best combinations

Hence  formula used is[tex]_n C_r=\frac{n !}{r ! (n-r) !}[/tex]

∴[tex]_n C_r=\frac{15!}{6!9!}[/tex]=[tex]\frac{15*14*13*13*12*11*10}{6*5*4*3*2}[/tex]=5,005 ways

B)

If order of choices is taken in considerations then one will use permutation to use

Hence using formula [tex]_{n} P_{r}=\frac{n !}{(n-r) !}[/tex]

=[tex]\frac{15!}{9!}[/tex]= 3,603,600 ways.

Learn more about permutation and combination https://brainly.com/question/16107928

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