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What does distinct permutations mean and how many distinct permutations can be formed from all the letters of word TOFFEE?

Gigili
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user1419170
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    Distinct permutations in this case means distinct $6$-letter "words" that can be made using each letter once and only once. TOFFEE is such a word, as is FEETOF. – André Nicolas Jul 26 '12 at 21:46

4 Answers4

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We know that the number of permutations of some given string of length $n$ is $n!$, however, we need to take into account the number of repeated permutations, we do this by counting the number of permutations of the repeated letters (in this case $F$ and $E$).

Therefore, we have:

$$\frac{6!}{2!^{2}}=180$$

Hope this helps!

Thomas Russell
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Here is another way to think about it. You have the word TOFFEE and six blanks 

- - - - - -

you want to place the letters in. Begin with the E. Choose two of the six blanks and pop the Es in. This can be done ${6\choose 2}$ ways. Now four blanks remain; place the Fs in these. There are ${4\choose 2}$ ways to do this. Two blanks remain for T and O; there are two ways to do this. So you get in toto, $$2{6\choose 2}{4\choose 2} = 2\cdot 15 \cdot 6 =180$$ ways to permute TOFFEE.

ncmathsadist
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The term "distinct permutations" takes into account that the word TOFFEE has two F's and two E's. This means that if we simply swap the two F's that the permutation is considered the same. You have to take this into account when doing the calculations for this problem.

Code-Guru
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TOFFEE

First consider that all the letters are distinct.

So 6!=720 possible permutations.
What's inside that 6! yo?

6!=6C2*2!*4C2*2!*2C2*2!
Let's explain it a little bit,

6C2*2! this part counts in how many ways, 2 positions can be selected from 6 positions
and 2 F's can be placed there(You could place 2 E's there too).
Now we have 4 positions left.If we place those 2 E's there number of permutation is 4C2*2!.For T and O other 2 positions so ,2C2*2!.

Now,Those 2! that we used for 2 F's and E's ain't right cause we have considered that all letters are distinct but they are not.

So out of 2! (FF and FF or EE and EE)combination only 1 combination(FF) is distinct and others are not.So instead of 2!, those counts should be 1
whice is,

6C2*1*4C2*1*2C2*2!=180 and tada that is same as 6!/(2!2!). That's why we divide by factorial of repeatation. Don't wanna divide ?
6!-number of repeated calculations
=6!- 6C2(2!-1)4C2(2!-1)*2C2*2! =180 Any way you want

Partha Dip
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