Number of blocks in this design $(V,\mathcal{B})$ $|V|=16$ and $\mathcal{B}$ has size $4$

Question

Let $(V,\mathcal{B})$ be a design in which $|V|=16$, each block in $\mathcal{B}$ has size $4$, and each pair of points occurs in precisely one block.

How many blocks are in this design?

Try 1: Now I can see that we have ${16 \choose 2}=120$ different blocks of size $2$. But this leaves the remaining two positions to be anything. They could be one of the other $2$-blocks.

I can't see how to get there from here. So I abandon this.

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Try 2: $16$ blocks with $1$ paired with the other numbers. Without recounting the $(1,2)$ pair $15$ with $2$. This is where I realise this can't work because I am ignoring the other two numbers

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Try 3:

every $1$ pairing

$\{1,2,3,4\},\{1,5,6,7\},\{1,8,9,10\},\{1,11,12,13\},\{1,14,15,16\}$

$2$

$\{2,5,6,7\},\{2,8,9,10\},\{2,11,12,13\},\{2,14,15,16\}$

$3$

$\{3,5,6,7\},\{3,8,9,10\},\{3,11,12,13\},\{3,14,15,16\}$

$4$

$\{4,5,6,7\},\{4,8,9,10\},\{4,11,12,13\},\{4,14,15,16\}$

$5$

$\{5,8,9,10\},\{5,11,12,13\},\{5,14,15,16\}$

$6$ has three, $7$ has three, $8$ has two, $9$ has two, $10$ has two and $11,12,13$ have one.

This gives a total of $5+4*3 + 3*3 + 2*3 + 3=35$

This might not even be right, and it is exhaustive. How can I do this properly???

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Try 4: I tried to construct a latin square, but soon realized this would only give me blocks of size $3$(I believe)

Answer 1

If every pair of elements is contained in precisely 1 block, then each element is contained in 5 blocks, because in each block there are 3 other elements, and $16=3\cdot 5+1$. We have 16 elements, therefore $16\cdot 5$ is 4 times the number of blocks. Thus the number of blocks must be 20, if there is such a design. From the catalogue of small designs we know, that there is precisely one such design.

Remark: your Try 3 does not work simply because you are putting some pairs of elements too many times together into one block. For example, pair 15,16 appears 5 times.

Answer 2

Your first attempt is a good start. You have ${16 \choose 2} = 120$ pairs that each need to show up precisely once together in a block. Each of the $b$ blocks contains ${4 \choose 2} = 6$ pairs, giving $b \cdot 6 = 120$. Thus $b = 20$.