Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to find the total cases where a four digit number has two like digits,three like digits, two pairs of like digits, four like digits etc...

share|improve this question
2  
Have you no thoughts of your own on how to do this? –  Gerry Myerson Aug 31 '12 at 12:09
1  
Four like digits is not hard, though the answer depends upon whether you allow 0000. –  Ross Millikan Aug 31 '12 at 13:05
add comment

2 Answers

I will assume that the four-digit numbers start with $1000$. If you allow things like $0075$ as four-digit numbers, then the calculations become substantially smoother.

I am sure that you can answer some of the questions easily. For example, one can list and count the "four like digit" numbers. They are $1111$, $2222$, and so on up to $9999$, a total of $9$.

Probably the next easiest to count is the "two pairs" case. The first digit can be chosen in $9$ ways. Suppose that for example we choose the first digit to be $4$. There are $3$ ways to decide where the other $4$ will go. And once we have decided that, there are $9$ choices for what digit the remaining two "holes" will be filled by. So we end up with $(9)(3)(9)$ "two pairs" numbers.

We now count the "one pair" numbers. It is maybe easiest to break up the work into two cases: (i) The pair involves the first digit and (ii) The pair does not involve the first digit.

For (i), the first digit can be chosen in $9$ ways. the location where the first digit is matched can then be chosen in $3$ ways. Once we have done that, we have two empty slots left. We can fill the leftmost one in $9$ ways. For each of these, we can fill the remaining empty slot in $8$ ways, for a total of $(9)(3)(9)(8)$. For (ii), the first digit can be chosen in $9$ ways. For each such choice, the slot that will not hold the pair can be chosen in $3$ ways, and the number that fills that slot can then be chosen in $9$ ways. Once we have done that, the number we have two of can be chosen in $8$ ways, for a total of $(9)(3)(9)(8)$. Thus the number of "one pair" numbers is $(9)(3)(9)(8)+(9)(3)(9)(8)$.

Finally, we deal with the "one triple" numbers. The triple could (i) Not involve the first digit or (ii) Involve the first digit. For (i), the first digit can be chosen in $9$ ways, and then the digit we have $3$ of can be chosen in $9$ ways, for a total of $(9)(9)$. For (ii), the first digit can be chosen in $9$ ways. For each such choice, the location of the odd digit can be chosen in $3$ ways, and then the odd digit can be chosen in $9$ ways, for a total of $(9)(3)(9)$. thus the number of "one triple" numbers is $(9)(9)+(9)(3)(9)$.

Remark: There are many other ways to organize the calculation. One clever thing to do would be to temporarily allow $0$ as an initial digit: then the calculations are smoother. After doing the computations, we subtract the forbidden cases. I would urge you to experiment with ways of counting. The numbers I have obtained can then serve as a check.

share|improve this answer
    
If the two pairs are distinct, then your second paragraph should read "And once we have decided that, there are 8 choices..." –  Byron Schmuland Aug 31 '12 at 15:37
    
@ByronSchmuland: I think it is $9$. Whatever choice we made for the first digit, there are $9$ choices of who will make up the other pair, since now $0$ is allowed. –  André Nicolas Aug 31 '12 at 15:42
    
Oh I see. The OP is not very clear on whether we wants to include four digit numbers like $0123$. You've decided to leave them out. –  Byron Schmuland Aug 31 '12 at 15:45
    
@ByronSchmuland: Thought I would go with the everyday meaning of four-digit number. –  André Nicolas Aug 31 '12 at 15:53
add comment

For this kind of counting argument, I like to use multinomial coefficients to keep things straight. Suppose I want to count the number of ordered strings of length four, taken from a set of ten digits, showing two different pairs; like this $0101$, $6446$, etc. Note that I am not going with the everyday meaning of four-digit number.

I first count the number of ways to select the digits, then multiply by the number of ways to arrange the digits.

$$ \begin{array}{cc} \text{Choose digits} & \text{Arrange them}\\[5pt] {10\choose 8,0,2}&{4\choose 2,2} \end{array} $$

This works out to $270$.

How do you read those multinomial coefficients? The first one gives the frequencies of the ten digits: there will be 8 digits that do not appear, 0 digits that appear exactly once, and 2 digits that appear twice. For the second one, under the four we have the pattern 2,2; that is, two pairs.

You can change the numbers to solve similar problems. Let's try 3 pairs in a string of length 6: $${10\choose 7,0,3}\times {6\choose 2,2,2}=10800. $$
How about one triple, one pair, and one single in a string of length 6: $${10\choose 7,1,1,1}\times{6\choose 1,2,3}=43200. $$

One last example: If you roll a die ten times, how many outcomes have one quintuple, two pairs, and a single? $${6\choose 2,1,2,0,0,1}\times {10\choose 1,2,2,5}=1360800. $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.