A counting problem that I came up with but cannot solve

[Random Variable]

For those who don't play golf, Taylormade makes a driver that allows you to change the clubs center of

gravity by relocating screws (made out of titanium and tungsten) in four locations in the head of the club. In

the base model, there are two 2-gram screws and two 10-gram screws. So obviosuly the numbers of ways

the screws can be arranged in the four locations is 4C2*2C2=4!/(2!2!)=6.


Taylormade also makes a professional version of the club that comes with two 2-gram screws, one 4-gram

screw, four 6-gram screws, one 8-gram screw, two 10-gram screws, one 12-gram screw, and one 14-gram

screw. Supposedly there are 883 ways those 12 screws can be arranged in the four locations in the clubs

head. I can't ever recall coming across such a problem in three semesters of probabilty and statistics (which

seems ridiculous). How would you solve such a problem? Everything I try I end up overcounting.

 

[her209]

Think of it this way.

AABCCCCDEEFG

How many different words can be made consisting of 4 letters?

 

[Random Variable]

Originally posted by: her209
Think of it this way.

AABCCCCDEEFG

How many different words can be made consisting of 4 letters?

Yeah, I know. But I don't think there is a quick way to solve such a problem.

 

[Random Variable]

Is this problem really that difficult?

 

[Spencer278]

Can't you do some thing like there is 12 screws that can go in the first hole, 11 that can go in the second, 10 in the 3rd, and 9 in the last. You then Have to cancle out any repeated combinations. SO you get
12 * 11 * 10 * 9 - some other number.,

 

[her209]

Originally posted by: Spencer278
Can't you do some thing like there is 12 screws that can go in the first hole, 11 that can go in the second, 10 in the 3rd, and 9 in the last. You then Have to cancle out any repeated combinations. SO you get
12 * 11 * 10 * 9 - some other number.,

Not all screws are distinct.

 

[TuxDave]

*bangs head on desk*

Dammit... why can't I get a solution?

 

[Random Variable]

Originally posted by: Spencer278
Can't you do some thing like there is 12 screws that can go in the first hole, 11 that can go in the second, 10 in the 3rd, and 9 in the last. You then Have to cancle out any repeated combinations. SO you get
12 * 11 * 10 * 9 - some other number.,

I was trying something like that. But it's confusing because there are so many things to subtract.

I think I'm just going to give up.

 

[Random Variable]

Originally posted by: her209

Originally posted by: Spencer278
Can't you do some thing like there is 12 screws that can go in the first hole, 11 that can go in the second, 10 in the 3rd, and 9 in the last. You then Have to cancle out any repeated combinations. SO you get
12 * 11 * 10 * 9 - some other number.,

Not all screws are distinct.

That's why he has the "- some other number."

 

[Spencer278]

Originally posted by: her209

Originally posted by: Spencer278
Can't you do some thing like there is 12 screws that can go in the first hole, 11 that can go in the second, 10 in the 3rd, and 9 in the last. You then Have to cancle out any repeated combinations. SO you get
12 * 11 * 10 * 9 - some other number.,

Not all screws are distinct.

That is why I have the - some other number. Find that other number is left to the reader.

 

[EyeMWing]

I can write a program to do it easily, but I don't know how you'd describe what it'd do in terms of conventional mathematics.

 

[CrackaLackaZe]

I dont understand the question

 

[CrackaLackaZe]

Oh you're looking for a formula to represent the number of combinations in the professional club, k.

 

[sniperruff]

my head just blew up just like a 1.4ghz thunderbird with an unproperly mounted HSF

 

[chuckywang]

I didn't get 883. I got 1423 as my answer.

1423 = "# of words with no repeated letters" + "# of words with two A's and no other repeated letters" + "# of words with two C's and no other repeated letters" + "# of words with two E's and no other repeated letters" + "# of words with three C's and no other repeated letters" + "# of words with four C's" + "# of words with two A's and two C's" + "# of words with two A's and two E's" + "# of words with two E's and two C's".

I tshink that should cover all the cases.

 

[her209]

Originally posted by: chuckywang
I didn't get 883. I got 1423 as my answer.

1423 = "# of words with no repeated letters" + "# of words with two A's and no other repeated letters" + "# of words with two C's and no other repeated letters" + "# of words with two E's and no other repeated letters" + "# of words with three C's and no other repeated letters" + "# of words with four C's" + "# of words with two A's and two C's" + "# of words with two A's and two E's" + "# of words with two E's and two C's".

I tshink that should cover all the cases.

Oops.. n/m.

 

[Random Variable]

.

 

[DrPizza]

I don't feel like finishing this, but I'll give you a good start....
(and I don't know if this is the simplest method)
First, since the order (or location) matters, I'd start it as a simple permutation...

So, the number of ways to place the screws is 12*11*10*9
Now, figure out how many ways there are such that you select only 1 2gram screw, because those repeat twice - subtract this number.

For this, since you have to have the 2 gram screw, (and only 1 2 gram screw), it can be in the first position, in which case you have 1*10*9*8 ways to pick the other screws or it could be in the second position...
So, 4*10*9*8 get's subtracted. Repeat for the 10 gram.

Similar, but slightly more complicated for subtracting for the 6 gram ones..

Now, some of the situations will be subtracted twice (sometimes you have a 2 gram and a 6 gram... you would have subtracted it for each case, so now you have to add these back in.)

Doesn't seem very simple to me, but probably do-able this way.

 

[DrPizza]

Or....... maybe easier... you can do it as several cases and add them together...
First, the # of ways without a 2, 6, or 10
Then, the # of ways with 1 2 and no 6 or 10
Then, the # of ways with 2 2's no 6 or 10
then, the # of ways with 1 2 and a 10
then, the # of ways with 2 2's and a 10
then, the # of ways with 2 2's and 2 10's
then... 1 2, 1 6
1 2, 2 6
1 2,3 6
2 2,1 6
2 2, 2 6
1 2, 1 6, 1 10
1 2, 2 6, 2 10
1 10, 1 6
1 10, 2 6
1 10,3 6
2 10, 1 6
2 10, 2 6
1 6
2 6
3 6
4 6

I think that's it! Most of those individual cases are quite simple.... granted it's a bunch to add up, but I can't think of a quicker way.

 

[her209]

Originally posted by: Vespasian
I get 940 that way.

7P4 + 6P2 + 6P2 + 6P2 + 6P1 + 1 + 1 + 1 +1 = 840+30+30+30+6+1+1+1+1 = 940

EDIT: Errr...that's not right. Give me a sec.

It should be:

7P4 + (6C2)(4!/2!) + (6C2)(4!/2!) + (6C2)(4!/2!) + (6C1)(4!/3!) + 4P4 + 4!/(2!2!) + 4!/(2!2!) + 4!/(2!2!)