What does distinct permutations mean and how many distinct permutations have the right to be created from all the letters of native TOFFEE?


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We understand that the number of permutations that some given string of size $n$ is $n!$, however, we should take into account the number of repeated permutations, we execute this by counting the number of permutations the the repetitive letters (in this case $F$ and also $E$).

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Therefore, we have:

$$frac6!2!^2=180$$

Hope this helps!


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Here is another way to think around it. You have actually the indigenous TOFFEE and six blanks

- - - - - -you desire to place the letter in. Begin with the E. Select two that the six blanks and pop the Es in. This can be excellent $6choose 2$ ways. Now 4 blanks remain; place the Fs in these. There room $4choose 2$ ways to execute this. Two blanks remain for T and O; there space two ways to carry out this. For this reason you obtain in toto, $$26choose 24choose 2 = 2cdot 15 cdot 6 =180$$ways come permute TOFFEE.


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The term "distinct permutations" takes right into account that words TOFFEE has two F"s and also two E"s. This method that if we just swap the two F"s that the permutation is thought about the same. You have to take this right into account as soon as doing the calculations because that this problem.


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TOFFEE

First think about that every the letters space distinct.

So 6!=720 feasible permutations.What"s inside the 6! yo?6!=6C2*2!*4C2*2!*2C2*2! Let"s define it a small bit,6C2*2! this component counts in how many ways, 2 positions have the right to be selected from 6 positionsand 2 F"s can be inserted there(You might place 2 E"s over there too).Now we have 4 location left.If we ar those 2 E"s therenumber the permutation is 4C2*2!.For T and O various other 2 positions so ,2C2*2!.

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Now,Those 2! that we offered for 2 F"s and also E"s ain"t right reason we have taken into consideration that every letters are distinct yet they are not.

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

6C2*1*4C2*1*2C2*2!=180 and also tada the is same as 6!/(2!2!). That"s why we divide by factorial the repeatation. Don"t wanna division ?6!-number of repeated calculations=6!- 6C2(2!-1)4C2(2!-1)*2C2*2!=180 Any method you want