Question

In a tennis tournament, every pair has to play with every other pair. Total 10 players are playing. Find the number of games played.

Answer 1

Hint: Use bijection rule to find the number of tennis matches played. Think about a bijective mapping between players playing tennis matches and making a group of two players. The latter can be calculated using combinations. Use the fact that the number of ways of selecting r people out of given n people is given by $^{n}{{C}_{r}}$.

Complete step-by-step solution -
Bijection rule: Consider two finite sets A and B. Let there exist a bijective mapping between elements of A and B. Then according to Bijection rule the number of elements in A is equal to the number of elements in set B, i.e. n(A) = n(B).
Consider A be the set of all tennis matches played, and B be the set containing all pairs of line segments with no common vertex formed by joining two points in a plane containing 10 non-collinear points.

Clearly, there exists a bijection between elements of set A and B.
If we form a group \left\\{ {{A}_{1}}{{A}_{2}},{{A}_{5}}{{A}_{6}} \right\\}, then the match is played between the pair \left\\{ \left\\{ {{P}_{1}},{{P}_{2}} \right\\},\left\\{ {{P}_{5}},{{P}_{6}} \right\\} \right\\} and vice versa.
Hence according to bijection rule, the number of elements in set A is equal to the number of elements in set B.

Calculation of n(B):
We first select two points which can be done in $^{10}{{C}_{2}}$ ways and from the remaining eight points, we select two points again, which can be done in $^{8}{{C}_{2}}$ ways.
Hence the total number of ways of forming the group in B is $^{10}{{C}_{2}}{{\times }^{8}}{{C}_{2}}$
But, every group is counted twice. Consider the case in which we select ${{A}_{1}}{{A}_{2}}$ first and then select ${{A}_{5}}{{A}_{6}}$. Now consider the case in which we select ${{A}_{5}}{{A}_{6}}$ first and then select ${{A}_{1}}{{A}_{2}}$. The group formed is the same but is counted twice.
Hence the total number of elements in B is $\dfrac{^{10}{{C}_{2}}{{\times }^{8}}{{C}_{2}}}{2}=\dfrac{\dfrac{10!}{8!2!}\times \dfrac{8!}{2!6!}}{2}=\dfrac{10!}{6!2!2!\times 2}=\dfrac{10\times 9\times 8\times 7}{8}=630$.
Hence by bijection rule, the number of elements in A is 630.
Hence the total number of test matches played is 630.

Note: Alternative method:
Select 4 players out of 10 players which can be done in $^{10}{{C}_{4}}$ ways.
Now from these selected four players, form two groups of two people each, which can be done in $\dfrac{4!}{2!2!2!}$ ways.
Hence the total number of tennis matches played $^{10}{{C}_{4}}\times \dfrac{4!}{2!2!2!}=630$.