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Question -

There are 10 professors and 20 students out of whom a committee of 2 professors and 3 students is to be formed. Find the number of ways in which this can be done. Further, find in how many of these committees:
(i) a particular professor is included.
(ii) a particular student is included.
(iii) a particular student is excluded.



Answer -

Given:

Total number ofprofessor = 10

Total number ofstudents = 20

Number of ways =(choosing 2 professors out of 10 professors) ├Ч (choosing 3 students out of 20students)

= (10C2)├Ч (20C3)

By using the formula,

nCr┬а= n!/r!(n тАУ r)!

10C2┬а├Ч┬а20C3┬а=10!/2!(10 тАУ 2)! ├Ч 20!/3!(20-3)!

= 10!/(2! 8!) ├Ч20!/(3! 17!)

= [10├Ч9├Ч8!]/(2! 8!) ├Ч[20├Ч19├Ч18├Ч17!]/(17! 3!)

= [10├Ч9]/2! ├Ч[20├Ч19├Ч18]/(3!)

= [10├Ч9]/(2├Ч1) ├Ч[20├Ч19├Ч18]/(3├Ч2├Ч1)

= 5├Ч9 ├Ч 10├Ч19├Ч6

= 45 ├Ч 1140

= 51300 ways

(i)┬аa particular professoris included.

Number of ways =(choosing 1 professor out of 9 professors) ├Ч (choosing 3 students out of 20students)

=┬а9C1┬а├Ч┬а20C3

By using the formula,

nCr┬а= n!/r!(n тАУ r)!

9C1┬а├Ч┬а20C3┬а=9!/1!(9 тАУ 1)! ├Ч 20!/3!(20-3)!

= 9!/(1! 8!) ├Ч 20!/(3!17!)

= [9├Ч8!]/(8!) ├Ч[20├Ч19├Ч18├Ч17!]/(17! 3!)

= 9 ├Ч [20├Ч19├Ч18]/(3!)

= 9├Ч[20├Ч19├Ч18]/(3├Ч2├Ч1)

= 9 ├Ч 10├Ч19├Ч6

= 10260 ways

(ii)┬аa particular studentis included.

Number of ways =(choosing 2 professors out of 10 professors) ├Ч (choosing 2 students out of 19students)

=┬а10C2┬а├Ч┬а19C2

By using the formula,

nCr┬а= n!/r!(n тАУ r)!

10C2┬а├Ч┬а19C2┬а=10!/2!(10 тАУ 2)! ├Ч 19!/2!(19-2)!

= 10!/(2! 8!) ├Ч19!/(2! 17!)

= [10├Ч9├Ч8!]/(2! 8!) ├Ч[19├Ч18├Ч17!]/(17! 2!)

= [10├Ч9]/2! ├Ч[19├Ч18]/(2!)

= [10├Ч9]/(2├Ч1) ├Ч[19├Ч18]/(2├Ч1)

= 5├Ч9 ├Ч 19├Ч9

= 45 ├Ч 171

= 7695 ways

(iii)┬аa particular studentis excluded.

Number of ways =(choosing 2 professors out of 10 professors) ├Ч (choosing 3 students out of 19students)

=┬а10C2┬а├Ч┬а19C3

By using the formula,

nCr┬а= n!/r!(n тАУ r)!

10C2┬а├Ч┬а19C3┬а=10!/2!(10 тАУ 2)! ├Ч 19!/3!(19-3)!

= 10!/(2! 8!) ├Ч19!/(3! 16!)

= [10├Ч9├Ч8!]/(2! 8!) ├Ч[19├Ч18├Ч17├Ч16!]/(16! 3!)

= [10├Ч9]/2! ├Ч[19├Ч18├Ч17]/(3!)

= [10├Ч9]/(2├Ч1) ├Ч[19├Ч18├Ч17]/(3├Ч2├Ч1)

= 5├Ч9 ├Ч 19├Ч3├Ч17

= 45 ├Ч 969

= 43605 ways

тИ┤┬аThe required no.of ways are 51300, 10260, 7695, 43605.

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