This problem involves finding the number of players in the intersection of two sets (Hockey and Cricket) within a larger group, given information about the total number of players and the sizes of various overlapping sets.
We use the Principle of Inclusion-Exclusion for three sets: H (Hockey), F (Football), and C (Cricket).
The formula is:
$|H \cup F \cup C| = |H| + |F| + |C| - |H \cap F| - |H \cap C| - |F \cap C| + |H \cap F \cap C|$Given values:
We need to find the number of players who play both Hockey and Cricket, denoted as $|H \cap C|$.
Substitute the known values into the Inclusion-Exclusion formula:
$44 = 26 + 24 + 24 - 8 - |H \cap C| - 12 + 5$Combine the known numbers:
$44 = (26 + 24 + 24 + 5) - (8 + 12) - |H \cap C|$ $44 = 79 - 20 - |H \cap C|$ $44 = 59 - |H \cap C|$Now, isolate $|H \cap C|$:
$|H \cap C| = 59 - 44$ $|H \cap C| = 15$Therefore, 15 players play both hockey and cricket.
Match List-I with List-II
| List-1 | List-II |
| (A) If X and Y are two sets such that n(X)= 17, n(Y)=23, n(X $\cup$ Y)=38, then n(X $\cap$ Y) is | (I) 20 |
| (B)) If n(X) = 28,n(Y) = 32,n(X$\cap$Y) = 10, then n(X$\cup$Y) is | (II) 10 |
| (C) If n(X) = 10, then n(7X) is | (III) 50 |
| (D) If n(Y) = 20, then n($\frac{Y}{2}$) is | (IV) 2 |
Choose the Correct answer from the options given below:
Consider the following relation R={(4,5),(5,4), (7,6),(6,7)} on set I={4,5,6,7}. Which of the following properties relation R does not have?
A. Reflexive property
B. Symmetric property
C. Transitive property
D. Antisymmetric property
Choose the correct answer from the options given below:
Find the least upper bound and greatest lower bound of $S=\{X,Y,Z\}$ if they exist, of the poset whose Hasse diagram is shown below: