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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 Mathematicians can not be included on the committee. 2. If 3 books are picked at random from a shelf containing 5 novels, 3 books of poems, and a dictionary, in how many ways this can be don if a) There is no restriction. b) The dictionary is selected? c) 2 novels and 1 book of poems are selected? 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 Mathematicians can not be included on the committee . 2. If 3 books are picked at random from a shelf containing 5 novels , 3 books of poems , and a dictionary , in how many ways this can be don if a ) There is no restriction . b ) The dictionary is selected ? c ) 2 novels and 1 book of poems are selected ?b​

Sagot :

Lanuel
  1. If there is no restriction, the number of ways is 350 ways.
  2. If one particular Statistician should be included, the number of ways is 150 ways.
  3. If two particular Mathematicians cannot be included on the committee, the number of ways is 105 ways.
  4. If there is no restriction, the number of ways this can be done is 84 ways.
  5. If the dictionary is selected, the number of ways this can be done is 28 ways.
  6. If 2 novels and 1 book of poems are selected, the number of ways this can be done is 30 ways.

How to determine the combination?

If there is no restriction, the number of ways in which a committee consisting of 2 Mathematician and 3 Statistician can be formed is given by:

Number of ways = ⁵C₂ × ⁷C₃

Number of ways = [5!/(2! × (5 - 2)!) × 7!/(3! × (7 - 3)!)]

Number of ways = [5!/(2! × 3!) × 7!/(3! × 4!)]

Number of ways = 20/2 × 210/6

Number of ways = 10 × 35

Number of ways = 350 ways.

If one particular Statistician should be included, the number of ways in which a committee consisting of 2 Mathematician and 3 Statistician can be formed is given by:

Number of ways = ⁵C₂ × ⁶C₂

Number of ways = [5!/(2! × (5 - 2)!) × 6!/(2! × (6 - 2)!)]

Number of ways = [5!/(2! × 3!) × 6!/(2! × 4!)]

Number of ways = 20/2 × 30/2

Number of ways = 10 × 15

Number of ways = 150 ways.

If two particular Mathematicians cannot be included on the committee, the number of ways in which a committee consisting of 2 Mathematician and 3 Statistician can be formed is given by:

Number of ways = ³C₂ × ⁷C₃

Number of ways = [3!/(2! × (3 - 2)!) × 7!/(3! × (7 - 3)!)]

Number of ways = [3!/(2!) × 7!/(3! × 4!)]

Number of ways = 6/2 × 210/6

Number of ways = 3 × 35

Number of ways = 105 ways.

Question 2:

If there is no restriction, the number of ways this can be done is given by:

Number of ways = ⁹C₃

Number of ways = [9!/(3! × (9 - 3)!)]

Number of ways = [9!/(3! × 6!)]

Number of ways = 84 ways.

If the dictionary is selected, the number of ways this can be done is given by:

Number of ways = ⁸C₂ × 1!

Number of ways = [8!/(2! × (8 - 2)!) × 1

Number of ways = [8!/(2! × 6!) × 1

Number of ways = 56/2 × 1

Number of ways = 28 × 1

Number of ways = 28 ways.

If 2 novels and 1 book of poems are selected, the number of ways this can be done is given by:

Number of ways = ⁵C₂ × ³C₁

Number of ways = [5!/(2! × (5 - 2)!) × 3!/(1! × (3 - 1)!)]

Number of ways = [5!/(2! × 3!) × 3!/(1! × 2!)]

Number of ways = 20/2 × 3

Number of ways = 10 × 3

Number of ways = 30 ways.

Read more on combination here: brainly.com/question/17139330

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