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

fjeihqo0 Apr 8, 2024

#1**+2 **

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.

hairyberry Apr 9, 2024

#1**+2 **

Best Answer

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.

hairyberry Apr 9, 2024