contestada

Suppose we want to choose 6 colors, without replacement, from 14 distinct colors. (a) How many ways can this be done, if the order of the choices matters? (b) How many ways can this be done, if the order of the choices does not matter?

Respuesta :

konrad509

a)

[tex]14\cdot13\cdot12\cdot11\cdot10\cdot9=2162160[/tex]

b)

just the divide the answer from a) by [tex]6![/tex]

[tex]\dfrac{2162160}{6!}=\dfrac{2162160}{720}=3003[/tex]

mathstudent55

Answer:

(a) 2,162,160

(b) 3,003

Step-by-step explanation:

(a) order matters

You can choose from 14 for the first pick. Then you have 13 left for the second pick. Then you have 12 left for the third pick. Keep going until you have 9 left for the 6th pick. The number when order matters is:

total = 14 * 13 * 12 * 11 * 10 * 9 = 2,162,160

(b) Order does not matter

Start with the same number as above for picking 6 out of 14. Since order does not matter, we divide by the number of ways you can arrange 6 items.

Since there are 6! ways of arranging 6 items,

total = 2,162,160/6! = 3,003

Otras preguntas