[SOLVED] probability question(digits)
A four-digit number, in the range 0000 to 9999 inclusive, is formed. Find the probability that
(a) the number begins or ends with 0,
(b) the number contains exactly two non-zero digits.
Can you help me solve this question? I have tried but didn't manage to get the real answer(but I strongly support my own answer).
Permutations and Combinations
You are raising two different points here. First, the difference between and .
Originally Posted by stpmmaths
- gives the number of ways of choosing objects from , and then arranging them in order. So, for instance, if we want to choose 2 from 4, and the objects are A, B, C and D, then the choice A, C will be counted as a different choice from C, A. There are therefore possible choices.
- gives the number of ways of choosing objects from , where the order doesn't matter. In the above example, A, C and C, A would be counted as the same choice, and there are therefore possible choices.
As an example, suppose that a committee chairman and secretary have to be chosen from a short-list of 4 people. Here, A, C is a different selection from C, A because A and C are fulfilling different roles. So you use . But if you have to choose 2 delegates to attend a conference, from 4 people, you'll use . Why? Because A, C and C, A are now the same choice - there's no difference between the roles of A and C.
So, in question (b), the number of different pairs of positions that can be occupied by a zero is , because one zero looks just like the other - the order in which you place them in the 'empty boxes' doesn't matter.
In each of the remaining boxes (containing non-zeros) the order in which we fill the boxes does matter: 0709 will be different from 0907, for instance. Each box can then be filled in 9 ways, so - using the 'r-s Principle' - the total number of ways in which these two boxes can be filled, one after the other, is .
Your second question is: how is this affected if we're not allowed to choose the same non-zero digit twice? The answer is this:
- The first stage - choosing which 'boxes' are to be filled with zeros - is unaltered. It's still 6 choices.
- In the second stage, suppose we fill the left-hand of the two remaining boxes first. There are 9 ways of doing this. Since we're not allowed to choose the same digit again, there are then 8 ways of filling the last box. Total number of choices: . Notice that this is the number of ways of choosing 2 items from 9, where the order matters; in other words, it's .
So the overall answer to your modified question would be choices, or a probability of .
Does that clear things up?