Q.The value of $n$, when $^{n}P_2 = 20$ is :

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 $n=8$: …

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.

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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.

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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

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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 …

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.

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