**Question:**

A survey of 50 students of a computer class gave the following data:

26 know PASCAL,

18 know FORTRAN,

30 know COBOL,

9 know both Pascal and FORTRAN,

16 know both Pascal and COBOL,

8 know both FORTRAN and COBOL,

43 know at least one of the three languages.

- From this we have to determine Marks:2+8

a. How many students know none of these languages?

b. How many students know all three languages?

** Solution: **n(U)=50 ; n(P)=26; n(F)=18 n(C)=30 ; n(P∩F)=9; n(P∩C)=16 n(F∩C)=8; n(P∪F∪C)=43

Where U = the set of students of computer class

P = the set of students who know PASCAL

F = the set of students who know FORTRAN

C = the set of students who know COBOL

a) By the difference rule, the number of the student who know none of these languages equal the total number of student minus the number who know at least one of the three languages

n(U) – n(P∪F∪C)=50-43=7

So 7 students know none of these languages.

b) n(P∩F∩C)=?

According to inclusion/exclusion Rule,

n(P∪F∪C)= n(P)+ n(F)+ n(C)- n(P∩F)- n(P∩C)- n(F∩C)+ n(P∪F∪C)

Put values

43=26+18+30-9-16-8+ n(P∪F∪C)

43=41+ n(P∪F∪C)

n(P∪F∪C)=43-41

n(P∪F∪C)=2

So 2 students know all three languages.

**Find the number of ways in which a teacher can select a team of 3 students from the class to work on a group project such that 1 student from each category.(that is one from those who knows PASCAL one from those who knows FORTRAN and one from those who knows COBOL). Marks:4**

Solution:

c(26,1)(18,1)(30,1)=26!/(26-1)!.1!*18!/(18-1)!.1!*30!/(30-1)!.1!

c(26,1)(18,1)(30,1)=26!/25!*18!/17!*30!/29!

c(26,1)(18,1)(30,1)=1012440

this is a combinatory problem in which repitition is not allowed and order doesnot matter

**Find the number of words that can be formed from the letters of the word FORTRAN. Marks:3 **

**Solution: **

**FORTRAN**

There are 7 letters of which two are R’s, hence required number of words are

Number of permutations = **7!/2!=2520**

**Find the number of words that can be formed from the letters of the word FORTRAN if the two R’s are to be next to each other.**

**Solution: **Consider the two R’s as one letter. Then there six letter which contain

Number of permutations =6!/1!=720

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