suppose I have a list of numbers [1,2,3....20].In how many ways can i select three numbers with repetitions from this list

Can somebody explain how to solve this question .I thought about it and it seems to me that there are 20 identical objects of first type ,20 identical objects of second and of third and we have to select three items from the three types which gives me $$\binom{20}{1} X \binom{20}{1}X \binom{20}{1}$$.Is this correct?

  • $\begingroup$ This would be correct. At each step, you have 20c1 = 20 choices. Then by multiplying the possibilities at each step (which you can do since the choices are independent of each other) as you have done gives you the answer. $\endgroup$
    – Azhao17
    Jul 6 '20 at 5:35
  • $\begingroup$ but i want the selections to be unique ie 1,19,1 should be counted same as 19,1,1.How do i do the calculation to include this $\endgroup$ Jul 6 '20 at 6:23
  • $\begingroup$ Ah I see. Do you count 5,6,7 and 7,6,5 as different? That is, does order matter here? $\endgroup$
    – Azhao17
    Jul 6 '20 at 6:28
  • $\begingroup$ no i count them as same $\endgroup$ Jul 6 '20 at 6:29
  • $\begingroup$ In this case you might consider breaking into cases: all different, 2 different numbers, all same. It would help here to use the choose function. $\endgroup$
    – Azhao17
    Jul 6 '20 at 6:33

If - as clarified in comments - the order does not matter the sequences differ only by counts of the corresponding items. This is equivalent to distributing 3 balls between 20 bins. The number of ways to perform this can be computed by stars and bars as: $$ \binom{3+20-1}3=\binom{22}3. $$

  • $\begingroup$ ah the answer is given same in textbook.Can u tell how this works.I know about stars and bars but i couldn't work out how to distribute the balls between the numbers $\endgroup$ Jul 6 '20 at 6:38
  • $\begingroup$ Have you already read the wikipedia page I linked? In my opinion it describes stars an bars in a very clear way. $\endgroup$
    – user
    Jul 6 '20 at 7:25
  • $\begingroup$ yeah i worked out how this works now.Thnaks $\endgroup$ Jul 6 '20 at 9:53

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