math help!?

Question

how many 3-digit counting numbers can be made using the digits 1,2,3,5,7, and 8?

a) with reptition?

b) without reptition?

Answer 1

a)
with reputation, this is how you do it:

You have 6 digits to choose from for the first number and since you can repeat digits, you have 6 digits for the 2nd and 3rd number as well. To get the total number of combinations, multiply these numbers:

# of combinations = 6*6*6 = 6^3 = 216 ways

b)
without repetition, this is a permutation problem. For the first number, you have 6 choices. For the 2nd number, you have 5 choices because you cannot choose the number chosen as the 1st one since you cannot repeat. For the 3rd number, you only have 4 choices since you cannot choose the other 2. Thus:

# of combinations = 6*5*4 = 120 ways

The general formula for a permutation is: n!/(n-k)! where,

n = # of choices
k = # of positions to place

In this example, n = 6 and k = 3. Plugging in you have:

6!/(6-3)! = 6!/3! = 6*5*4 = 120

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Hope this helps

Answer 2

a) with reptition?
6*5*4=120

b) without reptition?
6^3=216

Answer 3

Wow, combinatorics is so fun!

a) How many different possible first digits are there? 6. How many second digits? Since we allow repetition here, there are still 6 possible second digits. Also, there are six possible choices of third digit. Since each choice of three digits gives a different number, this gives 6*6*6 possible numbers.

b) I'll leave this one to you.

Answer 4

a) (10 + 3 - 1)! / ( 3! * (10-1)! = 220
b) 10! / (3! (10 -3)!) = 120