I have tried to solve this by excluding two % and performing permutation for non-distinct objects. As I excluded I left with 10 elements so n=10 and then considered non-distinct elements like # (n1=3),@(n2=2),$(n3=3),%(n4=2). n!/(n1!x n2!x n3!x n4!). Is this right approach?
Asked
Active
Viewed 96 times
3
-
Yes. That's correct. But please write the question title and description correctly and properly. – Dhruv Kohli Jun 15 '17 at 06:49
-
edited title to match second to last sentence of your description.. – Saketh Malyala Jun 15 '17 at 06:58
1 Answers
5
So what I'm assuming: $3$ hashtags, $2$ at's, $3$ dollar signs, $4$ percentage signs
You first fix two percentages, so you are arranging $3$ hashtags, $2$ at's, $3$ dollar signs, and $2$ percentages
This can be done in $\displaystyle \frac{(3+2+3+2)!}{3!2!3!2!}=\boxed{25200}$ ways.
Basically, arrange $10$ objects, and divide because each of the TYPES of objects are indistinguishable from one another.
Saketh Malyala
- 13,637