Permutation and combination when letters are repeated

[blamocur]

Please help me understand what am I missing. I was going to multiply 47040 by 2 and take that as an final answer.

I cannot help you without seeing a complete solution. BTW, I often find that explaining to others or writing down my solutions helps me to detect my own errors.

 

[mrtwhs]

First task is to line up M,A,A,A,A,S,T. As noted earlier this can be done in [imath]\dfrac{7!}{4!}[/imath] ways.

Now you can insert the 2H's and 2R's in four mutually exclusive ways: all four separate, 2 R's together and 2 H's separate, 2 H's together and 2 R's separate, 2 H's together and 2R's together. This yields

[imath]\dfrac{7!}{4!} \cdot \left( C(8,4) \cdot \dfrac{4!}{2!2!}+C(8,3)\cdot \dfrac{3!}{2!}+C(8,3)\cdot \dfrac{3!}{2!}+C(8,2)\cdot 2!\right)[/imath]

[imath]=210\cdot (70\cdot 6+56\cdot 3+56\cdot 3+28\cdot 2)=170,520[/imath]

 

[Maths Pro]

Thank you all for your inputs. It was indeed a great learning experience. If there are any more approaches I would like to know them as well. Finally I am now able to see how this problem works. Next challenge is circular permutations. But I will save it for another time.

 

[blamocur]

First task is to line up M,A,A,A,A,S,T. As noted earlier this can be done in [imath]\dfrac{7!}{4!}[/imath] ways.

Now you can insert the 2H's and 2R's in four mutually exclusive ways: all four separate, 2 R's together and 2 H's separate, 2 H's together and 2 R's separate, 2 H's together and 2R's together. This yields

[imath]\dfrac{7!}{4!} \cdot \left( C(8,4) \cdot \dfrac{4!}{2!2!}+C(8,3)\cdot \dfrac{3!}{2!}+C(8,3)\cdot \dfrac{3!}{2!}+C(8,2)\cdot 2!\right)[/imath]

[imath]=210\cdot (70\cdot 6+56\cdot 3+56\cdot 3+28\cdot 2)=170,520[/imath]

Nice! Definitely better than my solution.

 

[Dr.Peterson]

My method, adapted slightly from the QUEUES example, gives [math]\frac{1}{2}\left[\frac{8!}{4!}\cdot7\cdot8+\left(\frac{9!}{4!2!}-\frac{8!}{4!}\right)\cdot6\cdot7\right]=170,520[/math]
I arrange MSTAAAA with RR either together or separate, then insert the H's one at a time, and divide by 2.

 

[Maths Pro]

My method, adapted slightly from the QUEUES example, gives [math]\frac{1}{2}\left[\frac{8!}{4!}\cdot7\cdot8+\left(\frac{9!}{4!2!}-\frac{8!}{4!}\right)\cdot6\cdot7\right]=170,520[/math]
I arrange MSTAAAA with RR either together or separate, then insert the H's one at a time, and divide by 2.

Yes, I could get somewhere around the problem from your approach. Till then I was totally at sea.