Q.Find the number of distinct arrangements of the letters of the word 'STATISTICS'.

Ring (garland) arrangements of $n$ objects $=\dfrac{(n-1)!}{2}$; for $8$ flowers this is $\dfrac{7!}{2}$.

For a circular arrangement of $n$ objects, fixing one object to remove rotational repetition gives $(n-1)!$ arrangements. When the arrangement is a ring/garland (like a flower garland), a clockwise arrangement looks identical to its anticlockwise reflection, so we further divide by $2$:

$\text{Number of arrangements}=\frac{(n-1)!}{2}.$

For $n=8$:

$\frac{(8-1)!}{2}=\frac{7!}{2}.$

This circular-arrangement result is from the permutations section of the algebra chapter in the TN HSC Class-11 Business Mathematics syllabus.

The number of permutations of $n$ different things taken $r$ at a time with repetition allowed is $n^r$.

When repetition is not allowed, the count is $^nP_r=\dfrac{n!}{(n-r)!}$, because each choice reduces the pool by one.

When repetition is allowed, the pool never shrinks: the first place can be filled in $n$ ways, and so can the second, third, …, up to the $r$-th place, since a thing may be used again.

By the fundamental principle of counting:

$\underbrace{n\times n\times n\times\cdots\times n}_{r\text{ factors}} = n^r.$

When clockwise and anticlockwise are treated as the same, the number of circular permutations of $n$ objects is $\dfrac{(n-1)!}{2}$.

For $n$ distinct objects arranged in a circle, fixing one object to remove rotational repetition gives $(n-1)!$ distinct circular arrangements.

If, in addition, a clockwise arrangement and its mirror-image anticlockwise arrangement are regarded as the same (as with a garland of flowers or a bead necklace), every arrangement is counted twice, so we divide by $2$:

$\frac{(n-1)!}{2}.$

Write $^{n}P_2 = n(n-1)$, set it equal to $20$, and solve $n(n-1)=20$ to get $n=5$.

By the permutation formula,

$^{n}P_2 = \frac{n!}{(n-2)!} = n(n-1)$

So the condition becomes

$n(n-1) = 20 \;\Rightarrow\; n^2 - n - 20 = 0 \;\Rightarrow\; (n-5)(n+4)=0$

Thus $n=5$ or $n=-4$. Since $n$ must be a positive integer, $n=5$.

With repetition permitted each of the $r$ slots has all $n$ choices, so the number of arrangements is $n^{r}$.

Without repetition the count would be $^{n}P_r = \dfrac{n!}{(n-r)!}$, because each choice reduces the pool. But here repetition is allowed, so no place uses up an object:

• 1st place: $n$ ways
• 2nd place: $n$ ways
• $\vdots$
• $r$th place: $n$ ways

By the fundamental principle of counting the total is

$\underbrace{n\times n\times\cdots\times n}_{r\text{ times}} = n^{r}$

The leading digit has $9$ choices (it cannot be $0$), and the other $8$ places are filled from the remaining $9$ digits in ${}^{9}P_{8}=9!$ ways, so the total is $9\times 9!$.

We have $10$ digits available: $0,1,2,\dots,9$, and all $9$ digits of the number must be different.

First (leftmost) place: a 9-digit number must not begin with $0$, otherwise it would really be an 8-digit number. So the first place can be any of $\{1,2,\dots,9\}$ — that is $9$ ways.

Remaining $8$ places: one digit has already been used, so $9$ digits remain (this set now includes $0$). We arrange $8$ of these $9$ distinct digits in the $8$ places:

${}^{9}P_{8} = \frac{9!}{(9-8)!} = \frac{9!}{1!} = 9!.$

By the fundamental principle of counting, multiply the choices:

$\text{Total} = 9 \times 9!.$