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How many Binary Max-Heaps can be constructed from the elements {1,1,2,2,3,3,4,4} ?

for (1,1) there will be 1 tree

for (2,2,1,1) there will be 2 trees

for (3,3,2,2,1,1) there will be 6 trees

for (4,4,3,3,2,2,1,1) there will be (4,4,3,3,2,2,1,1)

(4,4,3,2,3,2,1,1),(4,4,3,3,1,-,-,-),(4,4,3,1,3,-,-,-),(4,4,3,2,1,-,-,-),(4,4,3,1,2,-,-,-),(4,4,3,2,2,-,-,-)

Now also in 2nd highest level there are also 3 combinations

(4,3,4,-,-,-,-,-)

(4,4,2,-,-,-,-,-)

So, I think it will be more than 44

@ma'am i think question asks for possibilities of heap considering all elements together.
im getting 40

Possible max heaps are 44

So as we have 4 nodes , those can be arranged in 4! ways but 2 similar nodes are their (2,2 and 1,1)

so total number of ways =4!/2!*2!   =  6 ways

Node 3 can be placed in 3 positions

For 2,1,1 we have 3 nodes(positions) after placing Node 3 and they can be arranged in 3!/2! ways because we have similar node 1

So totally we have 3* 3!/2! = 9 ways

If 1 is left child of 4 then other 1 must be child of node 1 as we cant place 2 or 3 in that position

So we have 3 positions

2,2,3 can be arranged in 3!/2!  = 3 ways

Total possibilities =6+9+3 =18 ways

All these above ways can also be for this heap

6+9+3 ways   =18 ways

In this we are left with 3 positions for 2,1,1 this can be arranged in 3!/2! =3 ways

Same as above In this we are left with 3 positions for 2,1,1 this can be arranged in 3!/2! =3 ways

generally as we have 3 positions and 2,1,1 is to placed we consider 3!/2! = 3 ways but 2 cant be child for 1 so we avoid this possibility then we are left with 2 ways

By adding all these possibilities we get 18+18+3+3+2 =44 ways

@sresta I tried to explain in detail you can check it now and let me know if anything wrong
I agree with u :)

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