SOLUTION: A certain college has 52 students participating in games.30 play netball, 25 play football , 10 play handball and football ,13 play netball and handball,9 play netball and football

Hi
30 - 9 -  13   =  8NB (Including the 2 all 3)
25 - 9 -  10   =  6FB (Including the 2 all 3)
32 play two 
52 - 32 - 8 - 6 = 6 HB only
Wish You the Best in your Studies.

For 3 subsets N, F and H, apply the inclusive-exclusive formula


    52 = N + F + H - NF - NH - FH + NFH.


Here NF, NH and FH denote in-pair intersections and NFH is the triple intersection.


In this formula, all the terms are given in the condition, except of H.


So we substitute known values and get this equation for H


    52 = 30 + 25 + H - 10 - 13 - 9 + 2,


simplify

    52 = 25 + H

and find 

    H = 52 - 25 = 27.


So, 27 play handball.


But we need only those who play handball and no other games.


So, we subtract NH= 13 and FH= 10  from  H= 27  and add  NFH= 2


        = 27 - 13 - 10 + 2 =   6.    ANSWER


ANSWER.  6 students play handball only.