Question 49867
<pre><font size =5><b>John constructed a dart board. The possible 
scores are 0, 1, 2 or 3. A score of 0 is 
obtained if the dart misses the board. Sally 
challenges John to a game consisting of each 
player throwing 6 darts at the board. The 
scores from the 6 throws are added. In how 
many ways can a total score of 15 or 16 be 
obtained?

<font color = "red">Note: The total score obtained from 1+0+0+3+3+2 
is considered different from the total score 
obtained from 0+0+1+3+3+2</font>

If it weren't for this note in red, then there
would be only these 3 basic ways to get 15:

 1. 0+3+3+3+3+3 = 15
 2. 1+2+3+3+3+3 = 15
 3. 2+2+2+3+3+3 = 15

and these 2 basic ways to get 16:

 4. 1+3+3+3+3+3 = 16
 5. 2+2+3+3+3+3 = 16

Without the red note, the answer
would be just 5. However, the note
tells us the numbers added may be 
arranged in any order, so for each 
of these 5 basic ways, we need to
apply the formula for the 
permutations of n things with 

r<sub>1</sub> maximum indistinguishable
things of one kind,
r<sub>2</sub> maximum indistinguishable
things of another kind,
···,
r<sub>k</sub> maximum indistinguishable
things of another kind.

This is n!/(r<sub>1</sub>!r<sub>2</sub>!···r<sub>k</sub>!)

In #1, the 5 3's are indistinguishable
Thus they can be rearranged 
6!/5! = 6 ways

In #2, the 4 3's are indistinguishable
Thus they can be rearranged 
6!/4! = 30 ways

In #3, the 3 2's and the 3 3's
are indistinguishable
Thus they can be rearranged 
6!/(3!3!) = 20 ways

In #4, the 5 3's
are indistinguishable
Thus they can be rearranged 
6!/5! = 6 ways

In #5, the 2 2's and the 4 3's
are indistinguishable
Thus they can be rearranged 
6!/(2!4!) = 15 ways

Thus the total number of ways of
getting 15 or 16 is

6+30+20+6+15 = 77 ways

Edwin</pre></b></font>