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.
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