Exercises: 1. Out of 5 Mathematician and 7 Statistician a committee consisting of 2 Mathematician and 3 Statistician is to be formed. In how many ways this can be done if a) There is no restriction b) One particular Statistician should be included c) Two particular Mathematicianscan not be included on the committee.
Answers
Step-by-step explanation:
Given Out of 5 Mathematician and 7 Statistician a committee consisting of 2 Mathematician and 3 Statistician is to be formed. In how many ways this can be done if a) There is no restriction b) One particular Statistician should be included c) Two particular Mathematicians cannot be included on the committee.
- So Out of 5 mathematicians, 2 mathematicians can be formed in 5C2 ways
- = 5! / 2! (5 – 2)!
- = 5 x 4 x 3 x 2 x 1 / 2 x 6
- = 10 ways
- Out of 7 statisticians 3 statisticians can be formed in 7C3 ways
- = 7! / 3! (7 – 3)!
- = 35 ways
- Therefore total number of possibilities will be 10 x 35
- = 350 ways
- Out of 5 mathematicians 2 mathematicians can be formed in 5 C2 ways.
- = 10 ways
- Out of 6 statisticians 2 additional statisticians can be formed in 6C2 ways
- = 6! / 2! (6 – 2)!
- = 30 / 2
- = 15 ways
- Therefore total number of possibilities will be 10 x 15
- = 150 ways
- Out of 3 mathematicians 2 mathematicians can be formed in 3C2 ways
- = 3! / 2! (3 – 2)!
- = 3 x 2 / 2 x 1
- = 3 ways
- Now out of 7 statisticians 3 statisticians can be formed in 7C3 ways
- = 7! / 3! (7 – 3)!
- = 5040 / 144
- = 35
- Therefore total number of possibilities will be 3 x 35
- = 105 ways
Reference link will be
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Answer:
(a) 350 ways (b) 150 ways (c) 105 ways.
Step-by-step explanation:
Given: Total 5 Mathematicians and 7 Statisticians, a committee of 2 Mathematicians and 3 Statisticians is to be formed.
To find: Number of ways if a) there is no restriction b) one particular Statistician should be included c) two particular Mathematicians cannot be included on the committee.
(a) No restriction
To form a committe of 2 Mathematicians and 3 Statisticians with restriction the number of ways are 5C2*7C3
= 5! / 2! (5 – 2)!*[7!/3!(7-3)!]
=10*35
=350 ways.
(b) one particular Statistician should be included
To form a committee of 2 Mathematicians and 3 Statisticians with one statistician fixed the number of ways are 5C2*6C2
We have taken 6C2 because one statistician is fixed so we have no choice for its selection we have to include that statistician
So, total ways=5C2*6C2
=5! / 2! (5 – 2)!*6!/2!(6-2)!
=10*15
=150 ways
(c) two particular Mathematicians cannot be included on the committee.
To form a committee of 2 Mathematicians and 3 Statisticians with two particular mathematicians exculded the number of ways are 3C2*7C3
We have taken 3C2 because two particular mathematicians always excluded so we have no choice for its selection.
Total ways=3C2*7C3
=3! / 2! (3 – 2)!*7! / 3! (7 – 3)!
=3*35
=105 ways
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