The number of words that can be made by permuting the letters of MATHEMATICS is
$1) 5040$
$2) 11!$
$3) 8!$
$4) 4989600$
First of all I do not understand the statement of the problem, I would like if some one tell me with an example.
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1Do you have any ideas? Can you show your work/speculations? – rurouniwallace May 14 '13 at 21:14
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@ZettaSuro I have no idea about what actually the problem saying me to find, If I knew I would have tried for sure. – Myshkin May 14 '13 at 21:15
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2@Tsotsi You're asked to find the cardinality of the set $$\mbox{${(x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}): (\forall k\in [11])(x_k\in {\mbox{A, E, I, C, H, M, S, T}} \mbox{ and A, M, and T occur exactly twice and the remaining occur exactly once.})}$}$$ – Git Gud May 14 '13 at 21:22
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Example: the number of words that can be made by permuting the letters of BEER is 12. The possibilities are EEBR, EERB, EBER, EREB, EBRE, ERBE, BERE, REBE, BREE, RBEE, BEER, REEB. – Daan Michiels May 14 '13 at 21:23
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okay so there may not be proper english words which is really found in dictionary – Myshkin May 14 '13 at 21:23
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@Tsotsi That is correct. – Git Gud May 14 '13 at 21:24
4 Answers
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To understand the question we assume a simpler word say BALL and see how many permutations of the word exist.
BALL can be permuted as ABLL, ALBL, ALLB, BALL, BLAL, BLLA, LABL, LALB, LBAL, LBLA, LLAB and LLBA. A total of $12$ possible permutations.
The permutations of a word is given by $$\frac{\text{(Total number of alphabets)}!}{\text{(Repetitions of A)}!\text{(Repetitions of B)}!\text{(Repetitions of C)}!...\text{(Repetitions of Z)}!}$$
The number of words that can be made by permuting the letters of MATHEMATICS is
$$\frac{11!}{2!2!2!}=4989600$$
Maazul
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$-1$ I did not want any answer, just what practical literal "meaning" of the statement like " you have to find these with an example", anyways thank you!but I have not understood your answer. there is still "why" – Myshkin May 14 '13 at 21:19
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I think "Total number of alphabets" would be better termed "total number of letters." – goblin GONE Mar 23 '17 at 23:19
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The problem is asking how many distinguishable ways you can rearrange the letters of the word MATHEMATICS. Let me give you a parallel example to see what is going on, here. Consider the word $$\text{MISSOURIANS}$$ let me rewrite it as $$\text{MIS}\color{blue}{\text{S}}\text{OUR}\color{red}{\text{I}}\text{AN}\color{green}{\text{S}}.$$ This has $11$ distinct objects in it, now that I've colored the repeated letters. There are $11!$ different ways we could arrange those ($11$ options for the first letter, then $10$ options for the second, then $9$ options for the third, and so on). Now, what if all of the letters letters $\text S$ were black, instead of different colors? Well, when they were different colors, each given arrangement of letters could be made into a completely different one simply by rearranging the $\text S$s (there are $3!$ ways to do this), but this makes no difference if they are all the same color. Hence, there are $$\frac{11!}{3!}$$ distinguishable ways to rearrange the letters of $$\text{MISSOUR}\color{red}{\text{I}}\text{ANS}.$$ From there, we see in a similar fashion that there are $$\frac{11!}{3!2!}$$ distinguishable ways to rearrange the letters of $$\text{MISSOURIANS}$$
See also this wikipedia article. Can you adapt this to your example?
Cameron Buie
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I know this has been answered with a result, but no one explained WHY the final formula is
$$\frac{n!}{n_1!n_2!...n_k!}$$
As described, we have 11 letters, which we divide into: 2 M's, 2 A's, 2 T's, 1 H, 1 E, 1 I, 1C, and 1 S
Next we calculate how many ways there are to place the M's, next A's, etc, and finally sum the outcomes: $$P= C(11,2)*C(9,2)*C(7,2)*C(5,1)*C(4,1)*C(3,1)*C(2,1)*C(1,1)$$ $$=\frac{11!}{2!9!}*\frac{9!}{2!7!}*\frac{7!}{2!5!}*\frac{5!}{1!4!}*\frac{4!}{1!3!}*\frac{3!}{1!2!}*\frac{2!}{1!1!}*\frac{1!}{1!}$$$$=\frac{11!}{2!2!2!1!1!1!1!1!} = 4989600 $$
I hope this helps for people in need in the future :)
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I am describing solution.MATHEMATICS contains 11 letters so they can be arranged by $11!$ ways but in this word there is some repeating letters M is 2 times,A is 2 times, T is 2 times.So in $11!$ ways there are some permutations presents which are repeat so we have to remove those permutations that's why divide the $11!$ by $2!\cdot 2!\cdot 2!$ (for 2 M,2 A and 2 T).suppose 1 M is on 5th place and 2nd M is on 6th place than if we swap their place the word is remain same.
iostream007
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