in a class of 42 students is play at least one of 3 game cricket hockey and football it is a found at 14 play cricket 20 play hockey 24 play football 3 play Both cricket and football to play both hockey and football and play all the three games find the number of student who play cricket but not only hockey
Answers
Answer:
Let the set of students who play cricket, hockey and football be C, H, F respectively.
Then,
n(C) = 14, n(H) = 20, n(F) = 24, n(C intersection F) = 3, n (H intersection F) = 2, n (C union H union F) = 42 and n (C intersection H intersection F) = 0.
We know:
n (C union H union F) = n(C) + n(H) + n(F) - n(C intersection H) - n(H intersection F) - n(C intersection F) + n (C intersection H intersection F)
Substituting the values, we get,
n(C intersection H) = 11
But, C = (C intersection H) intersection (C union H')
So, n (C) = n (C intersection H) + n (C intersection H')
[Since, (C intersection H) and (C intersection H') are disjoint sets]
Substituting the values, we get
n (C intersection H') = 3
Thus, the number of students who play cricket but not hockey is 3.
Step-by-step explanation:
Answer:
3
Step-by-step explanation:
Let the set of students who play cricket, hockey and football be C, H, F respectively.
Then,
n(C) = 14, n(H) = 20, n(F) = 24, n(C intersection F) = 3, n (H intersection F) = 2, n (C union H union F) = 42 and n (C intersection H intersection F) = 0.
Also,
n (C union H union F) = n(C) + n(H) + n(F) - n(C intersection H) - n(H intersection F) - n(C intersection F) + n (C intersection H intersection F)
Substituting the values, we get,
n(C intersection H) = 11
But, C = (C intersection H) intersection (C union H')
So, n (C) = n (C intersection H) + n (C intersection H')
[Since, (C intersection H) and (C intersection H') are disjoint sets]
Substituting the values, we get
n (C intersection H') = 3
Thus, the number of students who play cricket but not hockey is 3.
Other links:
https://brainly.in/question/7196394
https://brainly.in/question/5065639
#SPJ2