A hardware store sells numerals to people who make new houses but it has only got the numbers 3,5 and 8. How many different house numbers not more than 3 digits can be made from these numbers. The numbers can be repeated.
2006-10-30 20:26:10 · 6 answers · asked by Arnav 3 in Science & Mathematics ➔ Mathematics
When order matters (and order matters here because 35 is different from 53), the number of ways to combine n objects when you have k to choose from is k^n.
Here, we have three numbers to choose from (3, 5, and 8). So the number of 3-digit house numbers is 3^3 = 27, the number of 2-digit house numbers is 3^2 = 9, and the number of 1-digit house numbers is 3^1 = 3. So the answer is 27 + 9 + 3 = 39.
2006-10-30 20:30:55 · answer #1 · answered by Kevin L 2 · 2⤊ 0⤋
permutation of 3 item with repetitions
3^3
3*3*3
27
for number of possibilities for a 2-digit house number
3^2
3*3
9
for one digit house number
obviously there are only 3
so add them up together 3 + 9 + 27 = 39
2006-10-31 04:37:53 · answer #2 · answered by lazareh 2 · 0⤊ 0⤋
1 digit: 3 combos
2 digit: 3^2 combos
3 digit: 3^3 combos
total=3+9+27=39 possible combimations
2006-10-31 13:18:36 · answer #3 · answered by yupchagee 7 · 0⤊ 0⤋
Kevin is absolutely right.....39 different house numbers of 3-digits can be made.... Kudos!
2006-10-31 04:34:11 · answer #4 · answered by young_friend 5 · 0⤊ 0⤋
Assume 1st number is:358 and last can be=853
Therefore Numbers available with diff. combinations are=
854 - 358 = 496
2006-10-31 05:08:02 · answer #5 · answered by MY Regards to All 4 · 0⤊ 0⤋
Completely right.. couldn't have explained it better myself.
So i won't try :)
39 is correct though.
2006-10-31 04:53:26 · answer #6 · answered by CJ 3 · 0⤊ 0⤋