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PERMUTATIONS AND COMBINATIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSPERMUTATIONS AND COMBINATIONS1 Mark
Question
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
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Answer

Here total number of digits = 9
Number of digits used (no digit is repeated) = 3
$\therefore $ Number of permutations = $^9{P_3}$
$= \frac{{9!}}{{6!}} = \frac{{9 \times 8 \times 7 \times 6!}}{{6!}} = 504$

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