Answer:
100. 120
101. 28
102. 1
103. 24
Step-by-step explanation:
Number of Permutation (P) can be obtained as follow:
nPr = n! / (n– r)!
Number of Combination (C) can be obtained as follow:
nCr = n! / (n– r)! r!
With the above ideas in mind, let us determine the answers to the questions given above.
100. Evaluation of ₅P₄
nPr = n! / (n– r)!
₅P₄ = 5! /(5 – 4)!
₅P₄ = 5! / 1!
₅P₄ = 5 × 4 × 3 × 2 × 1 / 1
₅P₄ = 120
101. Evaluation of ₈C₂
nCr = n! / (n– r)! r!
₈C₂ = 8! / (8 – 2)!2!
₈C₂ = 8! / 6!2!
₈C₂ = 8 × 7 × 6!/ 6! × 2 × 1
₈C₂ = 8 × 7 / 2 × 1
₈C₂ = 56/2
₈C₂ = 28
102. Evaluation of ₇C₀
nCr = n! / (n– r)! r!
₇C₀ = 7! / (7 – 0)!0!
₇C₀ = 7! / 7!0!
Note: 0! = 1
₇C₀ = 7! / 7!
₇C₀ = 1
103. Evaluation 4!
4! = 4 × 3 × 2 × 1
4! = 24
