Question

Question 1 [Stars and Bars] In my pocket I never carry more than a dozen coins. I am talking about current Australian coins that come in six denominations: 5c, 10c, 20c and 50c (the ‘silver’ coins) and $1 and $2 (the ‘gold’ coins). Suppose I decided that I would carry a different collection of coins each day for as long as I could without ever repeating a collection. (Two collections are different if for at least one denomination there are more in one collection than in the other.)

(a) How many days could I keep this up if I have exactly twelve coins each day?

(b) How many days could I keep this up if I have no more than twelve coins each day? Hints: 1. There is no need to do any addition for this problem. 2. Sometimes I carry some worthless buttons with my coins.

(c) How many days could I keep this up if I have no more than twelve coins each day, but always have at least one gold coin?

(d) How many days could I keep this up if I have no more than twelve coins each day, but always have at least one coin of each denomination?

Answer 1

Answer 1: In my pocket I never carry more than a dozen coins. I am talking about current Australian coins that come in six denominations: 5c, 10c, 20c and 50c (the ‘silver’ coins) and $1 and $2 (the ‘gold’ coins). Suppose I decided that I would carry a different collection of coins each day for as long as I could without ever repeating a collection. (Two collections are different if for at least one denomination there are more in one collection than in the other.)

Solution:

5c, 10c, 20c and 50c (the ‘silver’ coins) and $1 and $2 (the ‘gold’ coins).

Suppose I decided that I would carry a different collection of coins each day for as long as I could without ever repeating a collection

(a) How many days could I keep this up if I have exactly twelve coins each day?

Exactly 12 coins each day

we have 6 different coins without repeating collection.

1) 6P1 = 6

2) 1/2*6P2 = 1/2 / 30 = 15

3) 1/3 * 6P3 = 1/3 * 120 = 40

4) 1/4 * 6P4 = 1/4 * 360 = 90

5) 1/5 * 6P5 = 1/5 * 720 = 144

6) 1/6 * 6P6 = 1/6 * 720 = 120

​​​​​​Therefore, total number of days = 6 + 15 + 40 + 90 + 144 + 120 = 415

Therefore, if I have exactly twelve coins each day I keep this up for 415 days.

(b) How many days could I keep this up if I have no more than twelve coins each day? Hints: 1. There is no need to do any addition for this problem. 2. Sometimes I carry some worthless buttons with my coins.

No more than 12 coins each day is equal to exactly 12 coins each day.

Therefore, if I have no more than 12 coins each day I keep this up for 415 days.

(c) How many days could I keep this up if I have no more than twelve coins each day, but always have at least one gold coin?

At least one gold coin. It may be $1 or $2 or both without repeating collection.

There are two possibilities:

1) Only single gold coin = 2

2) At least one gold coin = 9

For 6 coins follows the similar pattern.

(d) How many days could I keep this up if I have no more than twelve coins each day, but always have at least one coin of each denomination?

At least one coin of each denomination:

1/6 * 6P6 = 1/6 *720 = 120

Therefore, if I have no more than twelve coins each day, but always have at least one coin of each denomination then I keep this up for 120 days.