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| Question 1165301:  A survey of 250 SPC students was taken at registration. Of those surveyed:
 63 students had signed up for a Math course
 50 students had signed up for a Language Arts course
 21 students had signed up for both a Math and Language Arts course
 8 students had signed up for both a Math and Humanities course
 5 students had signed up for both a Language Arts and Humanities course
 1 students had signed up for all three courses
 2 students did not sign up for any of these classes
 How many students signed up for only Humanities (of these three)?
 Answer by ikleyn(52879)
      (Show Source): 
You can put this solution on YOUR website! . A survey of 250 SPC students was taken at registration. Of those surveyed:
 63 students had signed up for a Math course
 50 students had signed up for a Language Arts course
 21 students had signed up for both a Math and Language Arts course
 8 students had signed up for both a Math and Humanities course
 5 students had signed up for both a Language Arts and Humanities course
 1 students had signed up for all three courses
 2 students did not sign up for any of these classes
 How many students signed up for only Humanities (of these three)?
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We have an universal set U of 250 students
    and its subsets
M of 63 students;
L of 50 students;
H of unknown number of students;
   their in-pairs intersections
ML of 21 students;
MH of  8 students
LH of 5 students
    and finally their triple intersection
MHL of 1 student.
Based on this info, we can easily write an equation for the number of students who takes at least one of the three cources
    n(M U L U U) = m(M) + n(L) + n(H) - n(M ∩ L) - n(M ∩ H) - n(L ∩ H) + n(M ∩ L ∩ H). 
In this equation, the left side is 250-2 = 248.
In the right side, you know all the terms except of n(H), which I am going to find now.
Substituting all known terms, you get
    248 = 63 + 50 + n(H) - 21 - 8 - 5 + 1.
From this equation,
    n(H) = 248 - 63 - 50 + 21 + 8 + 5 - 1 = 168.
Now, to find n(H only), we should subtract n(MH) and n(LH) from n(H) and add n(MHL):
    n(H only) = 168 - 8 - 5 + 1 = 156.     ANSWER
Solved.
 
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