Answer to Question #96486 in Combinatorics | Number Theory for Isaac okocha

Question

50 of the Students were asked their subject combination 18 offered mathematics, 21.
offered Chemistry 16 offered biology& chemistry
offered mathematics and Chemisttry 8 students offered maths and biology
9 offered Chemistry and biology, While 5
offered the three sublect combination to
the venn diagram find
The numbed of student that passes mathematics
The number of studeats that affred Chemistry
but offered neither father mothe nor blology
The number of students that offered biology
but offered neither further maths nor Chemistry
The number of students who did not offer
any of the three Subject Combination.
​

Answer 1

Q96486

Solution:

As per the given question, we make a venn diagram and label each region with a variable. Here;

50 represents the total number of students.

M represents students offering Mathematics.

C represents students offering Chemistry.

B represents students offering Biology.

Next, we use the given conditions to form equations,

$a+b+c+d+e+f+g+h=50$ ------(1)

(total number of students)

$b+c+e+f=18$ -------(2)

(18 students offered M)

$c+d+f+g=21$ -------(3)

(21 students offered C)

$e+f+g+h=16$ -------(4)

(16 students offered B)

$e+f=8$ --------(5)

(8 students offered M and B)

$c+f=y$ --------(6)

(students offered M and C is not given in the question, so we assume it to be $y$, a known quantity)

$f+g=9$ --------(7)

(9 students offered B and C)

$f=5$ -------(8)

(5 students offering all three subjects)

Now, we solve them for the unknown variables.

$(7) and (8) \implies g=4$ -------(9)

(4 students offered B and C but not M)

$(6) and (8) \implies c=y-5$ -------(10)

($y-5$ students offered M and C but not B)

$(5) and (8) \implies e=3$ -------(11)

(3 students offered M and B but not C)

$Substituting (8),(9),(11) in (4)\implies h=4$ ----(12)

(4 students offered B but not M and C) (Answer)

$Substituting (8),(9),(10) in (3) \implies d=17-y$ ----(13)

($17-y$ students offered C but not M and B) (Answer)

$Substituting (8),(10),(11) in (2) \implies b=15-y$ ----(14)

($15-y$ students offered M but not C and B) (Answer)

$Substituting(8),(9),(10),(11),(12),(13),(14)$ $in (1) \implies a=y+7$

($y+7$ students offered none of the three subjects.) (Answer)