What is the number of arrangements of 5 different things taken 3 at a time in which 2 particular thing always occur?

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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