WT Chapter 4 in-action #3

[shweezy]

"A men's basketball league assigns every player a two-digit number for the back of his jersey. If the league uses only the digits 1-5, what is the maximum number of players that can join the league such that no player has a number with a repeated digit (e.g. 22), and no two players have the same number?"

I solved for this using an anagram of 12NNN, which yields 20 possibilities. However, I also subtracted 5 from this value, since the question states that no player can have repeated digits. Why is this wrong? Don't I have to account for 11, 22, 33, 44 and 55? THanks!

[tmmyc]

If every player has a two-digit number, which uses only digits 1-5, then:

First digit: 5 possible choices
Second digit: 5 possible choices
Total possible numbers: 5*5 = 25 possible choices

And here they are:

11
12
13
14
15

21
22
23
24
25

31
32
33
34
35

41
42
43
44
45

51
52
53
54
55

Subtract out those that have repeated digits (11, 22, 33, 44, 55) and you get

25 - 5 = 20 possibilities and hence 20 players.

Hope that helps.

[RonPurewal]

I solved for this using an anagram of 12NNN, which yields 20 possibilities. However, I also subtracted 5 from this value, since the question states that no player can have repeated digits. Why is this wrong? Don't I have to account for 11, 22, 33, 44 and 55? THanks!

you already excluded repeated digits, by using an anagram in the first place.

presumably, your anagram is meant to convey the idea that '1' is the first digit and that '2' is the second, so that, say, 'nn2n1' would represent jersey number 53. (correct me here if i'm wrong.)

in that case, then, repeated digits can't happen, because the '1' and the '2' can't fall in the same place. therefore, you don't have to make any additional corrections.

[Guest]

I solved for this using an anagram of 12NNN, which yields 20 possibilities. However, I also subtracted 5 from this value, since the question states that no player can have repeated digits. Why is this wrong? Don't I have to account for 11, 22, 33, 44 and 55? THanks!

you already excluded repeated digits, by using an anagram in the first place.

presumably, your anagram is meant to convey the idea that '1' is the first digit and that '2' is the second, so that, say, 'nn2n1' would represent jersey number 53. (correct me here if i'm wrong.)

in that case, then, repeated digits can't happen, because the '1' and the '2' can't fall in the same place. therefore, you don't have to make any additional corrections.

thanks ron and tmmyc. ron, actually 1 and 2 are just placeholders implying that the i care about order in two of the five spots (i.e. that this is a permuation as opposed to a combination, or YYNNN). with that said, i'm still not sure how numbers with the same digits are being accounted for in 12NNN?

[StaceyKoprince]

The anagram method already assumes that you cannot repeat or reuse whatever you're talking about - this is built in already so you don't have to worry about it. (And that also means you'd need to adjust this method for a problem on which you could repeat things - but that variation is rare enough that the basic formulas / techniques assume you CANNOT reuse items.)

Think about it this way: I've got 5 people from whom I'm choosing two to be on a team. Once I've picked Susie, I can't pick her again - there's only one Susie.

These five digits you have are the same thing (when using the anagram method): you've only got one of each. So once you pick, say, the digit 1, you've already used it up and can't use it again.