There are 4 ways to chose the first digit.
Then because there is to be no repetition, there are 3 ways to choose the second.
This means that there are 4 x 3 ways of setting down the first two digits
12 _ _
13 _ _
14 _ _
21 _ _
23 _ _
24 _ _
31 _ _
32 _ _
34 _ _
41 _ _
42 _ _
43 _ _
.... now you can take it from there.
# 230 # 203 # 320 # 302
With repetition, 94 = 6561. Without repetition, 9*8*7*6 = 3024.
9 odd numbers less than 100 can be formed. They are: 3,5,7,35,37,53,57,73 and 75.
There are 2000 such numbers.
36 two digit numbers can be formed...(:From Rafaelrz: The question can be stated as;how many permutations of two different digits can beobtained from a set of six different digits ?Answer:nPr equals n!/(n-r) ...... for n = 6, r = 26P2 equals 6!/(6-2)! equals 30 Permutations.
15 of them.
I take it that you want to make three digits numbers with 8,7,3, and 6 without repetition. The first digit cane be selected from among 4 digits, the second from 3 digits, the third digit from 2, hence the number of three digit numbers that can be formed without repetition is 4 x 3 x 2 = 24
Six (6)
9*8*7 = 504 of them.
3*2*1 = 6 of them.
64 if repetition is allowed.24 if repetition is not allowed.
# 230 # 203 # 320 # 302
1
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
125
There are 4*4*3*2*1 = 96 such numbers.
None :) You only gave us 4 numbers!