Statement 1; The  number  of  ways  of  choosing  15  couples  out  of   15  women  and  15 men  is 1240 and 
Statement  2;  The number  of   ordered  pairs  of   integers  (x,y) which satisfies  the  equation $y^2+6y+x^2=4$ is 8

Solution:    In  statement -1  the  required  number of  ways is $\frac{1}{15!}{(}^{15}{C}_1.{ }^{15}{C}_1)\left({(}^{14}{C}_1{.}^{14}{C}_1){(}^{13}{C}_1{.}^{13}{C_1}^{ })\mathrm{........}{(}^1{C_{1.}}^1{C}_1)\right)=\frac{1}{15!}{(15!)}^2=15!$
 
 In statement  -2  
$$\begin{gathered} y^2+6y+9=13-x^2 \\ \Rightarrow {(y+3)}^2=13-x^2 \end{gathered}$$

When  x=2                            y=0,-6
            X=-2                           y=0,-6
            X=3                             y=-1,-5
            X=-3                           y=-1,-5

 The number of order pair of integer is 8.