+0

# (a) We have three standard 6-sided dice, colored red, yellow, and green. In how many ways can we roll them to get a sum of 9? The dice are c

+5
1971
5
+1796

(a) We have three standard 6-sided dice, colored red, yellow, and green. In how many ways can we roll them to get a sum of 9? The dice are colored so that a red 2, yellow 3, and green 4 is a different roll from red 3, yellow 4, and green 2.

(b) We have 10 standard 6-sided dice, all different colors. In how many ways can we roll them to get a sum of 20?

Apr 30, 2015

#2
+17747
+10

Part b).

I believe that there are 23 different way to get a sum of 20.

Then we need to find the number of ways to get each of the 23 different ways.

For instance, one way is 6-6-1-1-1-1-1-1-1-1:  10! / (2! · 8!)   =  45 (different ways to get this result)

---   10 different dice are rolled, giving the 10! in the numerator

---   since the 6 is used twice, divide by 2!, since the 1 is used 8 times, divide by 8!

Etc., etc., etc.

May 2, 2015

#1
+17747
+15

Part a)

The different number combinations that will total 9:

1-2-6

1-3-5

1-4-4

2-2-5

2-3-4

3-3-3

Those that have 3 different values (like 1-2-6) have 6 different ways to get that combination.

Those that have 2 different values (like 1-4-4) have 3 different ways to get that combination.

Those that have only 1 way (like 3-3-3) have only 1 way to get that combination.

Thus, there will be 6 + 6 + 3 + 3 + 6 + 1 = 25 ways.

One way to determine the number of ways to get a particular combination is to use factorials.

The numerator is the factorial of the number of dice:  3!.

This numerator is divided by the factorial of each different number used -- for instance:

If the values used are 2-3-4:  3! / (1!·1!·1!) because each number is used once =  6.

If the values used are 2-2-5:  3! / (2!·1!) because the 2 is used twice and the 5 is used once = 3.

If the values used are 3-3-3:  3! / (3!) because the 3 is used three times.

May 2, 2015
#2
+17747
+10

Part b).

I believe that there are 23 different way to get a sum of 20.

Then we need to find the number of ways to get each of the 23 different ways.

For instance, one way is 6-6-1-1-1-1-1-1-1-1:  10! / (2! · 8!)   =  45 (different ways to get this result)

---   10 different dice are rolled, giving the 10! in the numerator

---   since the 6 is used twice, divide by 2!, since the 1 is used 8 times, divide by 8!

Etc., etc., etc.

geno3141 May 2, 2015
#5
+6
0

can you try to do this??? This is very complicated as you know

Robin+Jonathan  May 17, 2016
#3
-2

geno3141 -- I just wanted to point out that you missed 1-1-7 in counting your dice roles.

May 17, 2016
#4
+6
-1

THis, is wrong, as it only applies to DICE!!!!!!!!

Robin+Jonathan  May 17, 2016