Question

How many ways first 12 natural numbers ap?

Answer 1

The first question that came into my mind was "Minimum how many numbers make an AP?" and then I was convinced, by the definition of Arithmetic Progression which says that "an arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant", that the answer to the above mentioned question is 2 ,since there cannot be any concept called "difference between consecutive terms" unless there are at least 2 terms.

Getting back to the question, it says in how many ways can the first 12 natural numbers be divided into three different groups, such that the numbers in each group are in AP. Here goes the solution, step by step.

Case 1: Common difference=1

For example, any sequence like 10,11,12.. or 1,2,3...

So, we have to divide 12 numbers into 3 groups of consecutive numbers here, since the difference between two consecutive numbers is 1.

Applying combinatorics now, we can divide 12 numbers into 3 groups of consecutive numbers in following ways.

2,2,8               {1,2} {3,4} {5,6,7,8,9,10,11,12}2,3,7               {1,2} {3,4,5} {6,7,8,9,10,11,12}2,4,6               {1,2} {3,4,5,6} {7,8,9,10,11,12}2,5,5               {1,2 {3,4,5,6,7} {8,9,10,11,12}2,6,4               {1,2} {3,4,5,6,7,8} {9,10,11,12}2,7,3               {1,2} {3,4,5,6,7,8,9} {10,11,12}2,8,2               {1,2} {3,4,5,6,7,8,9,10} {11,12}3,2,7               {1,2,3} {4,5} {6,7,8,9,10,11,12}3,3,6               {1,2,3} {4,5,6} {7,8,9,10,11,12}3,4,5               {1,2,3} {4,5,6,7} {8,9,10,11,12}3,5,4               {1,2,3} {4,5,6,7,8} {9,10,11,12}3,6,3                     and so on like this3,7,24,2,64,3,54,4,44,5,34,6,25,2,55,3,45,4,35,5,26,2,46,3,36,4,27,2,37,3,28,2,2

So, we got 28 possible sequences.

Case 2: Common difference = 2

Now here comes some common sense. You are supposed to create 3 different groups, but then you took the common difference as 2. This means that you can create only two groups corresponding to common difference 2 and the other sequence should have some other common difference (and that common difference is obviously 1).

Here, the possible AP are

2,2,8           {1,3} {2,4} {5,6,7,8,9,10,11,12}3,3,6           {1,3,5} {2,4,6} {7,8,9,10,11,12}4,4,4           {1,3,5,7} {2,4,6,8} {9,10,11,12}5,5,2           {1,3,6,7,9} {2,4,6,8,10} {11,12}2,5,5           {1,2} {3,5,7,9,11} {4,6,8,10,12}6,3,3           {1,2,3,4,5,6} {7,9,11} {8,10,12}8,2,2           {1,2,3,4,5,6,7,8} {9,11} {10,12}

Case 3: Common difference = 3

The only case possible is {1,4,7,10} {2,5,8,11} {3,6,9,12}.

Hence, the total no. of ways are 28+7+1=36.

P.S: I have no idea why the answer mentioned in the book is 4. Either the answer is wrong (which surely is) or they have not well explained the need of the question.

Even if the groups were supposed to be equal, you see that out of these 36 combinations, the combination {4,4,4} appeared only 3 times.

Hope that was helpful.