Test Index

WAEC Further Mathematics 2022 Paper

Question : 18

Total: 45

Evaluate

4p_2 + 4C_2 - 4p_3

Solution:

Concept:

Use the standard formulas for permutation and combination to calculate each term separately.

Explanation:

The formula for permutation is

{}^nP_r = \frac{n!}{(n-r)!}

The formula for combination is

{}^nC_r = \frac{n!}{(n-r)! \, r!}

Here,

n = 4

First, calculate

{}^4P_2

{}^4P_2 = \frac{4!}{(4-2)!} = \frac{4!}{2!} = \frac{4 \times 3 \times 2!}{2!} = 12

Next, calculate

{}^4C_2

{}^4C_2 = \frac{4!}{(4-2)! \, 2!} = \frac{4!}{2! \, 2!} = \frac{4 \times 3 \times 2!}{2! \times 2!} = 6

Then, calculate

{}^4P_3

{}^4P_3 = \frac{4!}{(4-3)!} = \frac{4!}{1!} = 24

Now substitute these values into the given expression:

12 + 6 - 24 = -6

Answer:

-6

, which is option C.

Go to Question:

Prev Question

Next Question