Question

Using the digits 2 through 8, find the number of different 5-digit numbers such that: (a) Digits can be used more than once. (b) Digits cannot be repeated, but can come in any order. (c) Digits cannot be repeated and must be written in increasing order. (d) Which of the above counting questions is a combination and which is a permutation? Explain why this makes sense

Answer 1

a) There are 7 digits between 2 and 8, that are, 2,3,4,5,6,7,8. First case digits can be used more than once. We have to find a five digit number so that the digits can be repeated. So that each position can take any of 7 values. There are 5 such position. Therefore the total possible arrangements are .

b) Here also we have 7 digits where the digits cannot be repeated. Here the order of the arrangements are not important. Since the order is not factor we using the combination. We have to arrange 7 numbers in 5 positions. Therefore the total number of arrangements are .

c) Here we have to arrange the 7 digits in 5 position without repeating a single digit. Here the numbers should be in increasing order. The possible arrangements are 23456,23457,23458,23467,23468,23567,23568,23678,24567,24568,24678,25678,34567,34568,34678,35678,45678. Total number of arrangements are 17.

D) here the first two example is a combination problem since the order of arrangements is not a problem. Whereas in the case of third sample is an example of permutations since the order has an importance in this problem.