# Thread: [SOLVED] I have a problem.....

1. ## [SOLVED] I have a problem.....

It's a fairly straight forward one i think, but its been a long time since a studied maths!

The problem is...

There are 6 slots in a row.

One tab can be inserted into each slot.

There are 6 different coded tabs (e.g. 1, 2, 3, 4, 5, 6)

Therefore the total number of cominations is 6^6.

However, three or more tabs of the same code cannot lie next to each other.
i.e. the following combinations cannot exist:
2222222
655555
651113
144442
etc..... because there are 3 or more tabs of the same code next to each other.

What is the total number of allowable combinations?

2. For the first slot we have 6 possibilities.
For the second slot we have 5 possibilities (we can't put a tab with the same number as in the first position).
For the third slot we have 5 possibilities (we can\t put a tab with the same number as in the second slot)
..........................................
For the sixth slot we have 5 possibilities.
So, using the product rule of combinatorics, we have $6\cdot5\cdot5\cdot5\cdot5\cdot5=6\cdot5^5$ possibilities.

3. Originally Posted by red_dog
For the first slot we have 6 possibilities.
For the second slot we have 5 possibilities (we can't put a tab with the same number as in the first position).
You can have two the same adjacent, it is three that is forbidden.

For the third slot we have 5 possibilities (we can\t put a tab with the same number as in the second slot)
..........................................
For the sixth slot we have 5 possibilities.
So, using the product rule of combinatorics, we have $6\cdot5\cdot5\cdot5\cdot5\cdot5=6\cdot5^5$ possibilities.
RonL

4. Sorry...I misunterstood the problem. I'll think about it.

5. Count the number that look like: XXXYYY that is 30.
Count the number that look like: XXXYYZ that is $\left( 6 \right)\left( 5 \right)\left( 4 \right)\left( {\frac{{4!}}{2}} \right)$.
Count the number that look like: XXXYWZ that is $\left( 6 \right)\left( 5 \right)\left( 4 \right)\left( 3 \right)\left( {4} \right)$.
Thus, there are 2910 strings that have at least one triple.

Count the number that look like: XXXXYY that is $\left( 6 \right)\left( 5 \right)\left( {\frac{{3!}}{{2!}}} \right)$.
Count the number that look like: XXXXYZ that is $\left( 6 \right)\left( 5 \right)\left( 4 \right)\left( {3!} \right)$.
Thus, there are 810 with a string of four.

There are 60 with a string of five and 6 with a string of six.

Those are the one we do not want. Add up and subtract from $6^6$.

6. Originally Posted by spank
It's a fairly straight forward one i think, but its been a long time since a studied maths!

The problem is...

There are 6 slots in a row.

One tab can be inserted into each slot.

There are 6 different coded tabs (e.g. 1, 2, 3, 4, 5, 6)

Therefore the total number of cominations is 6^6.

However, three or more tabs of the same code cannot lie next to each other.
i.e. the following combinations cannot exist:
2222222
655555
651113
144442
etc..... because there are 3 or more tabs of the same code next to each other.

What is the total number of allowable combinations?
Hi spank .
I have a idea .

Case 1 : 3 XXX tags next to each other.
-Choosing 1 number : 10 choices.
-Choosing the tags's position : 4 choices. (XXX___,_XXX__,__XXX_,___XXX)
-Choosing the other tags : 9*8*7.

Case 2 : 4 XXXX tags next to each other.
-Choosing 1 number : 10 choices.
-Choosing the tags's position : 3 choices.
-Choosing the other tags : 9*8.

Case 3 : 5 XXXXX tags next to each other.
-Choosing 1 number : 10 choices.
-Choosing the tags' s position : 2 choices.
-Choosing the other tags : 9.

Case 4 : 6 tags next to each other.
-Choosing a number : 10.

So , the result is : 6^6 - (10*4*9*8*7 +10*3*9*8 +10*2*9 + 10)