Combinations and Permutations in Briefcase and Coin Problems

[lee534]

Hi all I need some assistance
1. Homework Statement with the attempt

How many 5-digit briefcase combinations contain

1. Two pairs of distinct digits and 1 other distinct digit. (e.g 12215)
I wasn't sure on which approach was correct.

10 * 9 * 8 (because there are three distinct digits)

or

10C2 * 5C2* 3C2 * 8 (because you have to take into account how the doubles can be orientated)

2. A pair and three other distinct digits. (e.g 27421)
same issue as above

10*9*8*7

or

(5C2*10) * 9 * 8 * 7

lastly
I have found this question before but I couldn't get an explanation:

How many ways are there to pick a collection of 12 coins from piles of pennies, nickels, dimes, quarters, and half-dollars? Base on the following condition:

1. there are only 10 coins in each pile.
16C4 - 5^2 because it's the total minus how many ways I can get from the 11th coin and the 12th coin. = 1795
2. There are only 10 coins in each pile and the pick must have at least one penny and two nickels?
1795 - 13C4(?)
my logic is that it's because 12-3+4 C 4 but I'm not sure if I have to set it to 12 or 10.

Thank you!

 

[kduna]

I only have a minute, but I'll see what I can get through:

1. First let's pick which digits are going to be used. There are 10 choices for the first paired digits, 9 choices for the second paired digits, and 8 choices for the lone digit. So there are $10*9*8$ ways to choose these digits. Now we have to order them. If we could distinguish our five numbers, the number of ways to order them would be $5!$. But we have those two indistinguishable pairs. So, for example, putting our first pair in the second and fourth position, we be counting this twice in the $5!$. So we need to divide by two for each of the pairs. Thus the number of ways to order our 5 numbers accounting for the fact that there are two pairs is: $\frac{5!}{2*2}$.

Thus the total number of combinations is: $10*9*8*\frac{5!}{2*2}$.

 

[lee534]

I only have a minute, but I'll see what I can get through:

1. First let's pick which digits are going to be used. There are 10 choices for the first paired digits, 9 choices for the second paired digits, and 8 choices for the lone digit. So there are $10*9*8$ ways to choose these digits. Now we have to order them. If we could distinguish our five numbers, the number of ways to order them would be $5!$. But we have those two indistinguishable pairs. So, for example, putting our first pair in the second and fourth position, we be counting this twice in the $5!$. So we need to divide by two for each of the pairs. Thus the number of ways to order our 5 numbers accounting for the fact that there are two pairs is: $\frac{5!}{2*2}$.

Thus the total number of combinations is: $10*9*8*\frac{5!}{2*2}$.

for the positioning

would it matter if the doubles were not next to each other for ex.

12125 or 13212?

 

[haruspex]

I only have a minute, but I'll see what I can get through:

1. First let's pick which digits are going to be used. There are 10 choices for the first paired digits, 9 choices for the second paired digits, and 8 choices for the lone digit. So there are $10*9*8$ ways to choose these digits. Now we have to order them. If we could distinguish our five numbers, the number of ways to order them would be $5!$. But we have those two indistinguishable pairs. So, for example, putting our first pair in the second and fourth position, we be counting this twice in the $5!$. So we need to divide by two for each of the pairs. Thus the number of ways to order our 5 numbers accounting for the fact that there are two pairs is: $\frac{5!}{2*2}$.

Thus the total number of combinations is: $10*9*8*\frac{5!}{2*2}$.

That's not quite right. E.g. 11223 gets counted both by picking two 1s, then two 2s and by picking them in the other order.

 

[lee534]

AH I solved it!, Thank you for replying to the posts!