Actually generating functions is used for the scenario where combination with limited repetition.But the given problem domain is different as the signal [1,1,2,2,2,2] is not the same as [2,2,2,2,1,1] .So it is the question of ordered partition rather..Hence we have to break the number 10 into groups of 1's and 2's and then find the no of arrangements for each group (partition) :
Case 1 :All 1's (in total 10 objects) :
So it can be done in 10! / 10! = 1 way as we have repeated 1's only.
Case 2 : 1 2 , 8 1's (in total 9 objects):
So this can be done in 9! / 8! = 9 ways [As repeptition of 8 ones , so needed to divide by 8!]
Case 3 : 2 2's , 6 1's :
This can be done in 8! / (6! * 2!) = (7 * 8) / 2 = 28 ways
Case 4 : 3 2's , 4 1's :
This can be done in 7! / (4! * 3!) = (5 * 6 * 7) / 6 = 35 ways
Case 5 : 4 2's , 2 1's :
This can be done in 6! / (4! * 2!) = 15 ways
Case 6 : All 2's :
This can be done in 5 ! / 5 ! = 1 ways
Hence total number of distinct signals = 1 + 9 + 28 + 35 + 15 + 1
= 89
Alternatively this question is similar to number of n - pennants which is nothing but nth Fibonacci number.
Therefore number of 10 - pennants = 10th Fibonacci Number = 89
