Welcome to our community

Be a part of something great, join today!

Problem of the week #23 - September 3rd, 2012

Status
Not open for further replies.
  • Thread starter
  • Admin
  • #1

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
MATH HELP BOARDS

is three words. The first word contains 4 letters, the second 4 letters and the third 6 letters. How many different ways can you group these letters into groups of 4, 4 and 6?

--------------------

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Admin
  • #2

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Congratulations to the following members for their correct solutions:

1) Sudharaka

Solution (from Sudharaka):

This problem is unfortunately ambiguous. I didn't specify if permutations of the same group count or not. If they do count, then here is Sudharaka's solution.
--------
Altogether there are 14 letters and there are two A's, two H's. All the other letters occur only once. If there are 14 different letters then there are \(14!\) different permutations. Let \(N\) be the number of permutations considering the repetitions of letters. For each permutation the two A's can be interchanged. Similarly the two H's can also be interchanged. Therefore,

\[2!\times 2!\times N=14!\]

\[\Rightarrow N=\frac{14!}{2!\times 2!}=21794572800\]

Finally each permutation above can be grouped into three groups of 4, 4, 6 letters. Hence the number of ways that the letters can be formed into the three groups of 4, 4, 6 letters is also, \(21794572800\).
--------
If permutations of the same group don't count, then the solution is:

\(\displaystyle \frac{ \binom{14}{4} \binom{10}{4} \binom{6}{6}}{2! 2!}\)
 
Status
Not open for further replies.