Please help me on this, thanks a lot.

In how many was can the letters of the word MURMUR be arranged without letting two letters which are the same be adjacent?

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- Dec 28th 2010, 06:37 PMrcshelp me on the arrangement please
Please help me on this, thanks a lot.

In how many was can the letters of the word MURMUR be arranged without letting two letters which are the same be adjacent? - Dec 29th 2010, 01:06 AMHallsofIvy
"MURMUR" has 6 letters so

**if**all the letters were distinct there would be 6! ways to arrange them. But there are two indistinguishable "M"s, two indistinguishable "U"s and two indistinguishable "R"s so there are distinct ways to arrange them.

If we treat both "M"s as a single letter, there would be 5 letters with two sets of pairs so ways to arrange the letters with the "M"s kept together. Similarly, there are 30 ways to arrange the letters with the "U"s kept together and 30 ways to arrange the letters witht the "R"s kept together. But the number of ways with the "U"s kept together**include**some with the "M"s and "R"s kept together.

There are ways to arrange the letters so that both "M"s and "U"s are kept together, 12 ways in which both "M"s and "R"s are kept together, and 12 ways in which both "U"s and "R"s are kept together. But each of those includes the 3!= 6 ways in which all "M"s, "U"s, and "R"s are kept together.

Altogether, there are 90 -30- 30- 30+ 12+ 12+ 12- 6= 30 ways to arrange the letters so that no pairs are together.

(Edited thanks to Wilmer.) - Dec 29th 2010, 01:30 AMWilmer
- Dec 29th 2010, 01:46 AMHallsofIvy
Yes, of course- 3!= 6.

- Dec 31st 2010, 02:50 AMrcs
thanks HallsofIvy. Great work!

- Dec 31st 2010, 11:24 AMArchie Meade
You can also take any letter and count the valid arrangements around it

Valid arrangements are

Valid arrangements are

Valid arrangements are

If there are 2 consecutive spaces, then there are 4 arrangements.

If there are 3 consecutive spaces, then there are 2 arrangements.

If there are 4 consecutive spaces, then there are 2 arrangements.

(2)

(4)

(2)

(2)

(4)

(4)

(2)

(4)

(4)

(2)

Total is