2
$\begingroup$

How would you go about calculating the number of permutations in ascending order.

Obviously if you had (a set of) 3 numbers you have $ 3! $ permutations:

(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1)

But only one of these is in ascending order (1,2,3).

Consider the lottery - picking 6 numbers from 49. Normally the order for this doesn't matter you just calculate '49 choose 6' but if the numbers have to be in ascending order how would you calculate that?

$\endgroup$
0

2 Answers 2

Reset to default
5
$\begingroup$

In the same way: The number of (strictly) ascending sequences of $6$ numbers chosen from $49$ is $\binom{49}{6}$. For as you have pointed out, once the $6$ numbers have been chosen, there is precisely one way to arrange them in ascending order.

$\endgroup$
4
  • $\begingroup$ Does this remain unchanged even if we have repeating numbers among the 6 chosen ones? $\endgroup$ Commented Aug 31, 2015 at 16:34
  • $\begingroup$ It changes. If we are looking for non-descending (weakly ascending) sequences, for $i=1$ to $49$ let $x_i$ be the number of times we use $i$. Then we want the number of solutions of $x_1+\cdots+x_{49}=6$ in non-negative integers. By Stars and Bars (please see Wikipedia) this number is $\binom{49+6-1}{6-1}$ or equivalently $\binom{49+6-1}{49}$. $\endgroup$ Commented Aug 31, 2015 at 16:45
  • $\begingroup$ The sum is $6$ because we are choosing $6$ (not necessarily different) numbers. Another way of saying it is that we are counting the number of $6$-element multisets. (I am mentioning the word to help you search). $\endgroup$ Commented Aug 31, 2015 at 17:01
  • $\begingroup$ You are welcome. $\endgroup$ Commented Aug 31, 2015 at 17:11
3
$\begingroup$

Hint: A combination is a permutation where order does not matter.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.