+51 Find the number of positive integers that satisfy both the following conditions:
Each digit is a 1 or a 3
The sum of the digits is 12
I got 3, its not correct. help is appreciated
+399 To find all possible combinations without ordering, we just set the number of 3s to be 1, 2, 3, or 4 and set the rest to 1s.
We can split into these following cases:
#1: 4 3s and 0 1s
#2: 3 3s and 3 1s
#3: 2 3s and 6 1s
#4: 1 3 and 9 1s
#5: 0 3s and 12 1s
Now we find the number of ways to order each case:
Case 1:
1 way to order this:
Case 2:
\(\frac{6!}{3!3!}=20\) ways to order this
Case 3:
\(\frac{8!}{6!2!}=28\) ways to order this
Case 4:
\(\frac{10!}{1!9!}=10\) ways to order this
Case 5:
1 way to order this.
Adding, 1+20+28+10+1 = 60 total numbers.
+399 To find all possible combinations without ordering, we just set the number of 3s to be 1, 2, 3, or 4 and set the rest to 1s.
We can split into these following cases:
#1: 4 3s and 0 1s
#2: 3 3s and 3 1s
#3: 2 3s and 6 1s
#4: 1 3 and 9 1s
#5: 0 3s and 12 1s
Now we find the number of ways to order each case:
Case 1:
1 way to order this:
Case 2:
\(\frac{6!}{3!3!}=20\) ways to order this
Case 3:
\(\frac{8!}{6!2!}=28\) ways to order this
Case 4:
\(\frac{10!}{1!9!}=10\) ways to order this
Case 5:
1 way to order this.
Adding, 1+20+28+10+1 = 60 total numbers.
