What did I do wrong in the permutations question.

Question

I was given the following question:

A hardware store sells numerals for house numbers. It has large quantities of the numerals 3, 5, and 8 but no other numerals. How many different house numbers, with no more than three digits can be made from these numbers?

So, I tried to solve it by multiplying 3 x 3 x 3 = 27 because there were 3 choices at each step. However, according answers of this book the answer is 39 and not 27.

Please explain what I did wrong.

Thank you :)

score 2 · Accepted Answer · answered Apr 16 '15 at 22:07

2

You've worked out the number of house numbers with exactly $3$ digits, but the question says `no more than' $3$ digits.

See if you can work out the number of house numbers with only $2$ or $1 $ digit!

answered Apr 16 '15 at 22:07

CameronJWhitehead

1,942

3

Oh. So that means I find 27 + 9 +3 = 39. Is that right? – anonymous Apr 16 '15 at 22:08
2

Yeah it is, well done! – CameronJWhitehead Apr 16 '15 at 22:09
1

Thank you for your support. – anonymous Apr 16 '15 at 22:09

What did I do wrong in the permutations question.

1 Answers1