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 $$\{A_1{A}_2,A_5{A}_6\}$$, then the match is played between the pair $\{\{P_1,P_2\},\{P_5,P_6\}\}$ 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 ${\displaystyle \frac{{}^{10}{C}_2{\times }^8{C}_2}{2}}={\displaystyle \frac{{\displaystyle \frac{10!}{8!2!}}\times {\displaystyle \frac{8!}{2!6!}}}{2}}={\displaystyle \frac{10!}{6!2!2!\times 2}}={\displaystyle \frac{10\times 9\times 8\times 7}{8}}=630$.
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 ${\displaystyle \frac{4!}{2!2!2!}}$ ways.
Hence the total number of tennis matches played ${}^{10}{C}_4\times {\displaystyle \frac{4!}{2!2!2!}}=630$.