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I know , there is formula to apply, s!(r-s+1)( n-s P k-s).

I tried with example, n=5 ; r=3 and s=2. I'm getting answer as 2!*(3-2+1)*3 = 2*2*3 = 12.

But, if I solve manually with example, consider n = {1,2,3,4,5} then possible arrangements are as follows:

145 154 245 254 345 354 415 425 435 451 452 453 514 524 534 541 542 543. Total count is 18. Where am I going wrong

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  • $\begingroup$ I'm referring this for formula: number 5 (b). books.google.co.in/… $\endgroup$ Aug 20, 2018 at 16:32
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    $\begingroup$ A suggestion... stop trying to search for formulas and trying to memorize different formulas for different situations. Instead, learn methods and strategies. This problem is easily approachable using the multiplication principle (a.k.a. rule of product). $\endgroup$
    – JMoravitz
    Aug 20, 2018 at 16:51
  • $\begingroup$ yea, thanks for the suggestion. Can you please help me out with this problem. $\endgroup$ Aug 20, 2018 at 16:57
  • $\begingroup$ The answers below already do a fine job of explaining the solution using multiplication principle. The general solution again being "pick what other numbers are used" followed by "pick how they are arranged." This can be done in $\binom{n-s}{r-s}\cdot r!$ ways, but again this formula should not be memorized since you can always just rederive it on the spot whenever it is needed. $\endgroup$
    – JMoravitz
    Aug 20, 2018 at 17:00
  • $\begingroup$ Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ Aug 20, 2018 at 17:15

3 Answers 3

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The formula is designed in a way such that it only accounts for permutations, where the chosen fixed elements appear together i.e. as a block. For example, in your case $451$ is allowed, whereas $415$ is not because $4$ and $5$ are the chosen elements. If you count only those, there are indeed $12$ of them.

The formula that deals with what you want is $r!{n-s \choose r-s}$. In this case, the number would be $3!{3 \choose 1}=18$, exactly what you got.

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  • $\begingroup$ You mean to say, 12 is when '4','5' are coming together? $\endgroup$ Aug 20, 2018 at 17:01
  • $\begingroup$ @RatnakarChinchkar Yeah. I meant exactly that. $\endgroup$ Aug 20, 2018 at 20:08
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I am not able to view your link, but here is how I would calculate it:

You are looking for your arrangement to contain both 4 and 5. So, choose one more number (from the remaining 3). Then permute the three numbers. My guess (and this is a guess) is that the formula you are looking at has something to do with multisets.

Anyway, choosing the third number, then permuting the three numbers would give you this calculation:

$$\dbinom{3}{1}3! = 3\cdot 3\cdot 2\cdot 1 = 18$$

which agrees with the number of arrangements you found by writing every one out.

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  • $\begingroup$ Interstellarprobe and SinTan , thank you guys. I'm trying to test formula : Number of permutation of n different things, taken r at a time, when s particular things are to be always included in each arrangement is , s! * (r-(s-1)) * n-s P r-s. In my case n = {1,2,3,4,5} and s= {4,5}. r=3. $\endgroup$ Aug 20, 2018 at 16:54
  • $\begingroup$ @Ratnakar Yes, if the 4 and the 5 must be adjacent, the answer is 12. Demonstration: $$\begin{matrix}145 \\ 154 \\ 245 \\ 254 \\ 345 \\ 354 \\ 451 \\ 541 \\ 452 \\ 542 \\ 453 \\ 543\end{matrix}$$ $\endgroup$ Aug 20, 2018 at 17:35
  • $\begingroup$ Yea, thanks a lot. $\endgroup$ Aug 20, 2018 at 17:43
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2 necessary items may be placed in 3P2 ways And remaining 1item out of 3 items in 3! Ways Hence answer =3P2*3P1=3*3!=18

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