Math Help - permutations

1. permutations

With the letters of the word MATEMATIK, how many 9 lettered words can be(meaningfull or meaningless) written starting and ending with a vowel?

My answer is 7.7!/2.2 but it seems to be wrong.

2. Re: permutations

Originally Posted by kastamonu
With the letters of the word MATEMATIK, how many 9 lettered words can be(meaningfull or meaningless) written starting and ending with a vowel?
My answer is 7.7!/2.2 but it seems to be wrong.
This is a messy case problem. Break it down into cases.
1) How many both begin and end with an a.
2) How many begin and with an a but end with either an e or i?
3) Double case 2). WHY?
4) How many neither begin nor end with an a?

3. Re: permutations

Many Thanks.I got the solution but can't understand the reason of the double case.

4. Re: permutations

Originally Posted by kastamonu
Many Thanks.I got the solution but can't understand the reason of the double case.
In case 2) it begins with a, case 3) is ends is a.

5. Re: permutations

I seperated others from ie because in IE middle term was different and there were AAs. In other cases there are two vovels in the middle term.

7. Re: permutations

Originally Posted by kastamonu
I seperated others from ie because in IE middle term was different and there were AAs. In other cases there are two vovels in the middle term.
1) How many both begin and end with an a.
$7!$

2) How many begin and with an a but end with either an e or i?
$2(7!) And that is the same for case 3) just start with the e or i. 4) How many neither begin nor end with an a?$\dfrac{2(7!)}{(2!)}=7!$ANSWER$6(7!)\$

8. Re: permutations

getting the same

9. Re: permutations

6(7!)/2!.2! according to my book.

10. Re: permutations

Originally Posted by kastamonu
6(7!)/2!.2! according to my book.
Actually that is correct, we forgot about the two M's and T's.

11. Re: permutations

Hi,
I understand where you got your original answer, but as pointed out, it is wrong. Here's one way to count. List all the possibilities for first and last letter and for each count the number of arrangements of the remaining 7 letters.
1. a..a: 7!/4 2 m's and 2 t's in the remaining letters
2. a..e: 7!/4 2 m's and 2 t's in the remaining letters
3. a..i: 7!/4 2 m's and 2 t's in the remaining letters
4. e..a: 7!/4 2 m's and 2 t's in the remaining letters
5. e..i: 7!/8 -- there are 2 a's, 2 m's and 2 t's in the remaining letters
6. i..e: 7!/8 -- there are 2 a's, 2 m's and 2 t's in the remaining letters
7. i..a: 7!/4 2 m's and 2 t's in the remaining letters

The sum of the arrangements 1 through 7 is then
$${6\cdot 7!\over 4}$$

12. Re: permutations

There are 9 letters with 2 A's, 2 T's, 2 M's

Vowels are A, A, E, I

The start and end can have AA, 2xAE, 2xAI, 2xEI

With AA 7!/{2! 2!] = 1260

With 2xAE = 2 x 7!/[2! 2!] = 2520

With 2xAI = 2 x 7!/[2! 2!] = 2520

With 2xEI = 2 x 7!/[2! 2! 2!] = 1260
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Total = 7560 <-----
Many Thanks. I also got the answer.

13. Re: permutations

Here is how I would do that: If we were to treat the two "A"s, two "M"s, and two "T"s as different (write one as " $A_1$" and the other " $A_2$" and similarly for the "M"s and "T"s) then there are 4 choices for the leading vowel. That leaves 3 choices for the final vowel. That leaves 7 letters that can be any where in the middle so 7! ways that can be done. That is there are (4)(3)(7!) ways to order the letters in " $M_1A_1T_1EM_2A_2T_2IK$".

But interchanging the "A"s does NOT give a new order. Treating the two "A"s as the same we have to halve that. Similarly for the two "M"s and two "T"s. The correct number is $\frac{(4)(3)(7!)}{(2)(2)(2)}= (3)(7!)/2= 7560$.

Many Thanks.