Permutations

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November 25th 2009, 05:03 PM
arze

Permutations

Find the number of permutations of four letters from the word MATHEMATICS.
There are 2 of M,A and T, So we have to divide the permutations by $(2!)^3$ . or do I find the number of permutations with two pairs of letters, and permutations with one pair of letters, and the number of permutations without any pairs of letters?
Thanks
November 25th 2009, 09:47 PM
aman_cc

Quote:

Originally Posted by arze

Find the number of permutations of four letters from the word MATHEMATICS.
There are 2 of M,A and T, So we have to divide the permutations by $(2!)^3$ . or do I find the number of permutations with two pairs of letters, and permutations with one pair of letters, and the number of permutations without any pairs of letters?
Thanks

I trust you will have to break it into cases where all four letter are different (like MATH) and 2 letters are same like (MMTH or MMTT etc) and then compute the permutations for each case and add them up
November 25th 2009, 10:50 PM
arze

I tried that but can't get the answer
for 2 pairs of letters, there are $3\times 2=6$ ?
for 1 pair of letters, there are $3\times 2\times 7=42$ and $3\times 5\times 4=60$ ?
and for no repetition $6\times 5\times 4\times 3=360$ ?
Thanks
November 25th 2009, 11:06 PM
aman_cc

Quote:

Originally Posted by arze

I tried that but can't get the answer
for 2 pairs of letters, there are $3\times 2=6$ ?
for 1 pair of letters, there are $3\times 2\times 7=42$ and $3\times 5\times 4=60$ ?
and for no repetition $6\times 5\times 4\times 3=360$ ?
Thanks

Can you details each case? For e.g. no repetition looks wrong to me - How do you get 360?
November 26th 2009, 12:03 AM
arze

oops wrong there, it should be 11-3=8, we have 1680. we take away the repeated letters right?
the one where two pairs is also wrong because the next question after this asks how many of the permutations contain to pairs of letters that are the same and the answer is 18, but i don't know how that is arrived at.
i'm going wrong somewhere here because the answer is supposed to be 2454.
November 26th 2009, 02:31 AM
aman_cc

Quote:

Originally Posted by arze

oops wrong there, it should be 11-3=8, we have 1680. we take away the repeated letters right?
the one where two pairs is also wrong because the next question after this asks how many of the permutations contain to pairs of letters that are the same and the answer is 18, but i don't know how that is arrived at.
i'm going wrong somewhere here because the answer is supposed to be 2454.

ok now. so let see how you did the following:

1. one pair
2. twp pair
November 26th 2009, 04:14 AM
arze

1. for two pairs i think it should be we have 3 choices then 2 right? so it would be 6.
2. for one pair we would have 3 choices first, then i don't exactly know what else. i think $3\times\frac{9!}{2!6!}$ .
November 26th 2009, 09:22 PM
aman_cc

Quote:

Originally Posted by arze

1. for two pairs i think it should be we have 3 choices then 2 right? so it would be 6.
2. for one pair we would have 3 choices first, then i don't exactly know what else. i think $3\times\frac{9!}{2!6!}$ .

For 2 pair: You have to select 2 letters from MAT. Once you have done that you need to find ways we can permute xxyy

So total ways - 3C2x(4!/(2!.2!)) = 3*6=24

So can you do 1 pair similarly now ?
November 26th 2009, 09:32 PM
arze

ok so for 2 pairs we will have 18 permutations.
for one pair, $^3C_1(^7C_3)(\frac{4!}{2!})=1260$
right?
Thanks
November 26th 2009, 09:55 PM
aman_cc

Quote:

Originally Posted by arze

ok so for 2 pairs we will have 18 permutations.
for one pair, $^3C_1(^7C_3)(\frac{4!}{2!})=1260$
right?
Thanks

For 2 pairs we will have 18 (correct) [I mis-calculated above]
For 1 pair however
we will have $^3C_1(^7C_2)(\frac{4!}{2!})$

Please note 7C2 - I'm sure u know the reason !