7 Sins

Consider Star and bars, three demons have to commit at least one sins, so work to do afterward is to distribute four sins between three people. It is equivalent to putting two bars in between four balls, $\dbinom{(4+(3-1)}{(3-1)}$ , which is 15

Sins are identical, and at least one sin is committed by each. So the problem comes down to the number of combinations of 4 (7-3) of identical balls (the remaining 4 sins) in 3 bins (the three sinners). The solutions are 4 in one bin : 3C1=3 cases 3 in one bin, 1 in a second bin, 0 in a third bin: 6 cases 2 in one bin, 1 in a second bin, 1 in a third bin: 3cases 2 in one bin, 2 in a second bin, 0 ina third bin: 3 cases

Total: 3+6+3+3=15

Since each demon needs to commit at least one sin, let D1(Demo 1) commits ( $a+1$ ) sins, D2 commits ( $b+1$ ) sins and D3 commits ( $c+1$ ) sins.

$a+1$ + $b+1$ + $c+1$ = 7, where $a$ , $b$ and $c$ are all bigger or equal to 0.

This gives us $a$ + $b$ + $c$ = 4.

According to the star and bars techniques (i.e., $*|**|*$ is an example in this case), we have $6 \choose 2$ = 15 ways!!

Abe 1 1 1 1 1 Raith 1 2 3 4 5 Ghost 5 4 3 2 1

If at least 1 sin is to be committed by all the 3 then this list will be repeated 3 times by keeping no of sins committed by the Abe or Raith or Ghost 1 in a list. Number of different ways they can split up the work is 3*5=15.

Classic stars and bars question.

Because the sins can be thought of us $7$ identical objects in a row imagine these as our stars:

$*******$

But each of the sinners (Abe, Raith and Ghost) must commit at least one sin. Remove three stars:

$****$

Now, we want to figure out how many ways we can divide the remaining sins among the sinners. Think of the sinners as $3$ bins we want to put sins into, and we want to count the ways to fill those bins (a bin may have up to $4$ sins or $0$ sins). Thus the number of bars we need to divide our stars into bins is $3-1=2$ . This fact can be extrapolated to other stars and bars questions, the number $n$ of individuals, bins, boxes, etc. always requires $n-1$ dividers to divide the number of items we are distributing.

Now, our answer is

$4 + (3-1) \choose 3-1$

Assume the sins are balls. Now imagine you have to get a 2 sticks between the 7 balls. that is between the six gaps that are formed in the 7 balls. that would split the seven sins balls into three parts. Gotcha. 6C2 and you have the answer

Put one apple (sin) in their baskets first, then they will have at least one apple

Distribute the remaining 4 apples to them is just like putting 2 separating bars and 4 apples in 6 slots

example

The no. of ways to put the 2 bars into 6 slots $= 6C2 = 15$

Since each sin as good as any other, they can be considered identical. We find all the triples where the elements add up to 7.