Counting Problems

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October 13th 2009, 03:11 PM
melody

Counting Problems

A full set of tiles for the game Tantrix contains 56 hexagonal tiles. Each tile has three dierent coloured lines on it, from four possible colours: red, yellow, green and blue. There are 14 tiles with no blue lines on them, 14 with no red and so on.

To commence a game, a player selects 6 tiles from the set of 56.

i. In how many ways can the player select 6 tiles?
ii. In how many ways can the player select 6 tiles with no blue on them?
iii. In how many ways can the player select 6 tiles, each with a blue line?

This is a game, please explain to me how to do it?
October 13th 2009, 07:40 PM
utopiaNow

Quote:

Originally Posted by melody

A full set of tiles for the game Tantrix contains 56 hexagonal tiles. Each tile has three dierent coloured lines on it, from four possible colours: red, yellow, green and blue. There are 14 tiles with no blue lines on them, 14 with no red and so on.

To commence a game, a player selects 6 tiles from the set of 56.

i. In how many ways can the player select 6 tiles?
ii. In how many ways can the player select 6 tiles with no blue on them?
iii. In how many ways can the player select 6 tiles, each with a blue line?

This is a game, please explain to me how to do it?

i. 56 ways to chose the first tile * 55 (since we already chose 1) ways to choose the 2nd. etc. to get:
56*55*54*53*52*51 = answer, or 56P6 if you know nPr notation and meaning.

ii. There are only 14 with no blue, therefore same method as i. except now we only have a total of 14 to choose from. Do you understand how to do this one?

iii. Same method as i. and ii. except the total now is 56 - 14 = 42. 42 tiles have a blue line, since 14 out of 56 don't. So no our total is 42, so how many ways to choose 6 from 42? Follow the same method I outlined in i. except in i. the total we started with was 56.
October 13th 2009, 08:05 PM
melody

Quote:

Originally Posted by utopiaNow

i. 56 ways to chose the first tile * 55 (since we already chose 1) ways to choose the 2nd. etc. to get:
56*55*54*53*52*51 = answer, or 56P6 if you know nPr notation and meaning.

ii. There are only 14 with no blue, therefore same method as i. except now we only have a total of 14 to choose from. Do you understand how to do this one?

iii. Same method as i. and ii. except the total now is 56 - 14 = 42. 42 tiles have a blue line, since 14 out of 56 don't. So no our total is 42, so how many ways to choose 6 from 42? Follow the same method I outlined in i. except in i. the total we started with was 56.

ii. so 14*13*12*11*10*9 = 2162160 ways or 14P6 = same previous answer right?
iii. so 42*41*40*39*38*37 = 3776965920 or 42P6 = same previous answer right?
October 13th 2009, 08:22 PM
utopiaNow

Quote:

Originally Posted by melody

ii. so 14*13*12*11*10*9 = 2162160 ways or 14P6 = same previous answer right?
iii. so 42*41*40*39*38*37 = 3776965920 or 42P6 = same previous answer right?

Correct. Make sure you understand why nPr or $\frac{n!}{(n-r)!}$ , where $n$ is the total number of elements available and $r$ is the number of elements to be selected, gives you the correct answer.

You might want to refer to: Permutation
October 14th 2009, 01:58 PM
awkward

Just a comment on the solutions posted thus far: They assume the order of the tiles matters, so, for example, the order in which the first 6 tiles is drawn is significant.

If order doesn't matter, then you have a different problem and a different solution.