In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
It is given that 9 courses are available from which 2 courses are compulsory.
Since 2 courses are compulsory, thus the subjects remained are 7 and also from the 5 courses that are to be chosen has 2 courses compulsory which results in 3 subjects that are to be arranged then the number of ways of selecting them are the combination of 7 courses taken3 at a time.
The formula for the combination is defined as,
C n r = n ! ( n − r ) ! r ! .
Substitute 7 for n and 3 for r in the above formula.
C 7 3 = 7 ! ( 7 − 3 ) ! 3 ! = 7 ! 4 ! 3 !
Cancel the common factors by factorizing the bigger term to the factorial.
The formula to calculate the factors of a factorial in terms of factorial itself is,
n ! = n ( n − 1 ) ! n ! = n ( n − 1 ) ( n − 2 ) ! [ n ≥ 2 ] ⋮
The combination is written as,
C 7 3 = 7 × 6 × 5 × 4 ! 4 ! 3 × 2 × 1 = 7 × 6 × 5 3 × 2 × 1 = 35
Thus, the number of ways that the courses are selected is 35.
It is given that 9 courses are available from which 2 courses are compulsory.
Since 2 courses are compulsory, thus the subjects remained are 7 and also from the 5 courses that are to be chosen has 2 courses compulsory which results in 3 subjects that are to be arranged then the number of ways of selecting them are the combination of 7 courses taken3 at a time.
The formula for the combination is defined as,
Substitute 7 for n and 3 for r in the above formula.
Cancel the common factors by factorizing the bigger term to the factorial.
The formula to calculate the factors of a factorial in terms of factorial itself is,
The combination is written as,
Thus, the number of ways that the courses are selected is 35.
