$n$
balls numbered $1,2,\ldots,n$
that you are placing in $n$
distinct buckets. In how many ways can you do this such that: $\textit{(i)}$
no bucket remains empty? $\textit{(ii)}$
exactly one bucket remains empty?

 You have a bag of
$13$
sweets, each of different flavours. How many combinations of$5$
sweets can you form?  A mixed lacrosse team is to consist of
$6$
boys and$6$
girls. In how many ways can this team be assembled given the coach has$8$
boys and$10$
girls to choose from? $10$
athletes run a race, but one particular athlete had an accident and will be last on the scoreboard. How many possible rankings are there?
 You have a bag of

For part
$\textit{(i)},$
how many balls should be placed in each bucket? 
For part
$\textit{(ii)},$
what is the maximum number of balls in a bucket? 
How many buckets must contain the maximum number of balls?

How many combinations of balls are there for the bucket containing
$2$
balls? 
How many possible choices are there for the bucket containing
$2$
balls? 
Once
$2$
balls are placed into an arbitrary bucket, what is the distributions (or configuration) for the remaining balls in the remaining buckets? 
How many ways are there to distribute
$(n2)$
balls in$(n1)$
buckets, with at most$1$
ball per bucket? 
Consider the first ball of the remaining
$(n2)$
balls that you place. How many buckets are there to choose from? 
After the above first 3 balls are placed, what about the number of possible choices for the next ball?

$\textit{(i)}$
All buckets must each contain one ball. For the first bucket, there are$n$
balls that we may choose from. In the second bucket, there are$(n1),$
and so on. We therefore get$n!$
permutations.$\textit{(ii)}$
The condition is equivalent to exactly one bucket having$2$
balls, one bucket having$0$
balls, and the rest having$1$
ball each. Consider the bucket containing
$2$
balls. There are$n$
choices for this bucket, and$\binom{n}{2}$
possible combinations to select 2 out of$n$
balls.  We are then left with
$(n2)$
balls to distribute between$(n1)$
buckets, with no more than$1$
ball per bucket. For our first ball, there are$(n1)$
buckets to choose from. For our second ball, there are$(n2)$
buckets to choose from, and so on until the last ball, which has$2$
buckets as possible choices.  This leaves us with
$1$
empty bucket, as required.
Multiplying, we get
$\binom{n}{2} \cdot n \cdot (n1)!= \binom{n}{2}n!.$
 Consider the bucket containing