Calculate the numbers of ways to arrange 6 letters out of 7 letters of

  • Thread starter Thread starter xender_c
  • Start date Start date
  • Tags Tags
    Numbers
AI Thread Summary
To calculate the number of ways to arrange 6 letters from the 7 letters in "SETTEES," the result is 210. The process involves considering different cases based on the repetition of letters, such as having three identical letters of one kind, two of another, and one of a third. Using the multinomial theorem simplifies the calculation by allowing for the counting of choices at each stage and then dividing out equivalent arrangements. Specifically, arrangements of letters with duplicates require accounting for indistinguishable letters to avoid overcounting. Ultimately, the method involves multiplying the permutations of distinct arrangements and dividing by the factorial of the counts of identical letters.
xender_c
Messages
1
Reaction score
0
Calculate the numbers of ways to arrange 6 letters out of 7 letters of the word `SETTEES'.
Answer:210
Could anyone explain this to me ?thx
 
Physics news on Phys.org


try making different cases, for example 3 alike of 1 kind 2 alike of another kind and 1 of third kind.
or if you have read multinomial theorem try using that.it is simpler.
 


I find it most helpful to go through the mental process of actually choosing an arrangement and counting the choice one makes at each stage. You multiply the numbers for each stage to get the total multiplicity of choices. You then consider if you created any equivalent arrangements and divide those out. Example: Arrangements of two of the letters "ADD".

First I would distinguish the two D's { A, D1, D2}.
There are three letters and I have 3 choices for the first in my arrangement. There are two remaining and I have 2 choices for the 2nd. That's 3 x 2 = 6 choices.

Now since we can't really distinguish D1 vs D2 we divide out the ways we may have multiply chosen the same arrangement.

Of those 6 arrangements starting with any of the 2 permutations of D1 and D2 will give me an identical sequence of choices. So I divide by 2!=2 6/2 = 3.
AD DA and DD.

In your case there are 3!=6 ways to permute the three E's times 2!=2 ways to permute the S's and 2!=2 ways to permute the T's. That's 6x2x2=24 equivalent starting configurations. Figure the way to arrange 6 of 7 distinct letters and then divide by 24.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top