Let A, B and C be the sets such that A∪B=A∪C and A∩B=A∩C. Show that B = C
Or
A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58 men and only three men got medals in all three sports? How many received medals in exactly two of the three sports?
We have A∪B=A∪C
⇒(A∪B)∩C=(A∪C)∩C
→(A∩C)∪(B∩C)=C
⇒ [∵(A∪C)∩C=and using distributive law (A∪B)∩C=(A∩C)∪(B∩C)]
⇒(A∩B)∪(B∩C)=C [∵A∩C=A∩B] …(i)
Again, A∪B=A∪C
⇒(A∪B)∩B=(A∪D)∩B
⇒B=(A∩B)∪(C∩B)
⇒ [∵(A∪B)∩B=B and using distributive law (A∪C)∩B=(A∩B)∪(C∩B)]
⇒B=(A∩B)∪(B∩C) …(ii)
From Eqs. (i) and (ii), we get.
B = C
or
Let F denotes the set of men who received medals in football, B denotes the set of men who received medals in basketball and C denotes the set of men who received medals in cricket.
Then, we have
n(F)=38,n(B)=15.n(C)=20
n(F∪B∪c)=58 and n(F∩B∩C)=3
Now, n(F∪B∪C)=n(F)+n(B)+n(C)−n(F∩B)−n(8∩C)−n(F∩C)+n(F∩B∩C)
⇒58=38+15+20−n(F∩B)−n(B∩C)−n(F∩C)+3
⇒n(F∩B)+n(B∩C)+n(F∩C)=76−58=18
Now, number of men who received medals in exactly two of the three sports
=n(F∩B)+n(B∩C)+n(F∩C)−3n(F∩B∩C)=18−3×3=18−9=9
Thus, 9 men received medals in exactly two of the three sports.