SOLUTION: How many permutations of 5 letters each in examination

Algebra ->  Permutations -> SOLUTION: How many permutations of 5 letters each in examination      Log On


   



Question 825296: How many permutations of 5 letters each in examination
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
In the word "examination",

there are two each of letters 'a', 'i', and 'n', which are indistinguishable,
and there is one each of letters 'e', 'o', 'm', 't', and 'x'.

Case 1.  Those permutations which contain a maximum of 1 letter from
{a,e,i,m,n,o,t,x} which is P(8,5).  That's P(8,5) permutations.
-----------------------------------------------------------------
The following three cases 2,3,and 4 yield the same number of permutations 
each:

Case 2.  Those which contain two a's, and a maximum of one 'i' or 'n'
We choose the positions for the two a's C(5,2) ways and the remaining 3
from {e,i,m,n,o,t,x} in P(7,3) ways.  That's C(5,2)P(7,3) permutations.

Case 3.  Those which contain two i's, and a maximum of one 'a' or 'n'
We choose the positions for the two i's C(5,2) ways and the remaining 3
from {a,e,m,n,o,t,x} in P(7,3) ways. That's C(5,2)P(7,3) permutations.

Case 4.  Those which contain two n's, and a maximum of one 'a' or 'i'
We choose the positions for the two n's C(5,2) ways and the remaining 3
from {a,e,i,m,o,t,x} in P(7,3) ways. That's C(5,2)P(7,3) permutations.
-----------------------------------------------------------------------
The following three cases 5,6, and 7 yield the same number of permutations
each:

Case 5.  Those which contain two a's and two i's, and a maximum of one 'n'
We choose the positions for the two a's C(5,2), and the positions for the
two i's C(3,2), and one letter from {e,m,n,o,t,x} in P(6,1) ways. That's
C(5,2)C(3,2)P(6,1) ways.

Case 6.  Those which contain two a's and two n's, but a maximum of one i
We choose the positions for the two a's C(5,2), and the positions for the
two n's C(3,2), and one letter from {e,i,m,o,t,x} in P(6,1) ways. That's
C(5,2)C(3,2)P(6,1) ways.

Case 7.  Those which contain two i's and two n's, but a maximum of one 'a'
We choose the positions for the two i's C(5,2), and the positions for the
two n's C(3,2), and one letter from {a,e,m,o,t,x} in P(6,1) ways. That's
C(5,2)C(3,2)P(6,1) ways.


Total = P(8,5) + 3·C(5,2)P(7,3) + 3·C(5,2)C(3,2)P(6,1) = 

6720 + 3·10·210 + 3·10·3·6 = 6720 + 6300 + 540 = 13560.

Answer: 13,560

Edwin