How many different three digit numbers can be formed using the integers 1 to 6 if no integer occurs twice in a number?
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Correct Answer: Option B
Explanation:
This is equivalent to
\(P_{3}=\frac{6 !}{(6-3)}=\frac{6 \times 5 \times 4 \times 3 !}{3 !}=120\)
This is equivalent to
\(P_{3}=\frac{6 !}{(6-3)}=\frac{6 \times 5 \times 4 \times 3 !}{3 !}=120\)