# permutation

• January 8th 2010, 01:10 AM
thereddevils
permutation
(1) Digital devices store information in the form of digits using only the digits 0 or 1 . Calculate the number of codes which can be stored if they consist 10 digits .

(2) Calculate the number of ways to arrange 6 letters out of the 7 letters of the word SETTEES .
• January 8th 2010, 08:16 AM
Drexel28
Quote:

Originally Posted by thereddevils
(1) Digital devices store information in the form of digits using only the digits 0 or 1 . Calculate the number of codes which can be stored if they consist 10 digits .

(2) Calculate the number of ways to arrange 6 letters out of the 7 letters of the word SETTEES .

What have you tried? Is this combination or permutation? Is it true that the first one is just $_{10} P_2$??
• January 8th 2010, 08:23 AM
Shanks
(1)each digit has two options:0,1. therefore the number of codes can be stored is $2^{10}$.
(2)Classification:
case (a): only one S in the 6 letters
then the 6 letters are one S, two T, three E. By the multiplication principle, the numbers of arrangements: ${6\choose 1}{5\choose 2}$.

case (b): two S in the 6 letters
then the 6 letters are possibly (bi): two S, one T, three E; (bii): two S, two T, Two E.
for case(bi): the number of arrangements: ${6\choose 1}{5\choose 2}$.
for case(bii): the number of arrangements: ${6\choose 2}{4\choose 2}$.
Sum these numbers up to get the total number of arrangements.
• January 8th 2010, 08:46 AM
All 10 digits of the code can be 0 or 1.
That's 2 possibilities for every digit....

2(2)(2)(2)... 10 twos multiplied together, which is ?

6 of the 7 letters of SETTEES has 3 possibilities due to the 3 distinct letters.

2 of all 3 letters,
3 E's, 2 S's and a T,
3 E's, 2 T's and an S,
1 E is not possible as we would only have 5 letters.

Since we cannot tell the difference between the arrangements of the three E's unless we colour them for instance, then

3!(n)=6! where n=the number of distinguishable arrangements if all the other letters were distinct.
They are not, however.

Along with that, we also cannot distinguish between the two T's, or the two S's.
Hence 3!2!n=6! finds the number of arrangements with 3 E's in both cases.
2!2!2!m=6! finds the number with 2 of each.

Therefore $2n+m=2\frac{6!}{3!2!}+\frac{6!}{2!2!2!}$
• January 8th 2010, 08:55 AM
Shanks
We get the same anser since we do the same things.
• January 8th 2010, 09:00 AM
Hi Shanks,

Yep,
i'd been typing this up and noticed you had posted
and thought, "will i or won't i?"
at least they are a little different but your way is nicely structured.
• January 8th 2010, 06:34 PM
thereddevils
Quote:

Originally Posted by Shanks
(1)each digit has two options:0,1. therefore the number of codes can be stored is $2^{10}$.
(2)Classification:
case (a): only one S in the 6 letters
then the 6 letters are one S, two T, three E. By the multiplication principle, the numbers of arrangements: ${6\choose 1}{5\choose 2}$.

case (b): two S in the 6 letters
then the 6 letters are possibly (bi): two S, one T, three E; (bii): two S, two T, Two E.
for case(bi): the number of arrangements: ${6\choose 1}{5\choose 2}$.
for case(bii): the number of arrangements: ${6\choose 2}{4\choose 2}$.
Sum these numbers up to get the total number of arrangements.

Thanks Shanks , that helps , hey you are from Beijing , my grandpa was from Guang Dong , nice meeting you .