# possibilities for entry code of a building [closed]

Exercice True or Flase

The entry code in a building consists of four digits As this building is located in Paris, some occupants want the selected code contains the number "$75$". Then there would be $168$ different possible codes.

My thoughts:

since the code is going to be :

$$75xy\quad x75y\quad \text{or} \quad xy75 \quad \text{with}\quad x,y \in \{0,1,2,3,4,5,6,7,8,9 \}$$ then

• $75xy$ gives us $8\times8$ possibitlies
• $x75y$ gives us $10\times 8$ possibitlies
• $xy75$ gives us $10\times 8$ possibitlies

thus there as possibilities $8\times8+10\times 8+10\times 8=224$

## closed as unclear what you're asking by BruceET, Tim Raczkowski, graydad, user91500, EmptySep 10 '15 at 6:27

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• "As this building is located in Paris"????? – barak manos Sep 9 '15 at 19:28
• Probably the exercise intends the four digit to be all different. In that case you have $7\times8\times3=168$ possible codes. – Aretino Sep 9 '15 at 19:35
• Why aren't your three cases equally numerous? Did you neglect to say first digit can't be 0? Confusing all around. Why assume four different digits? Please edit. BTW: Paris is French "Department" 75. Also 'French 75' is (deservedly obscure) cocktail. – BruceET Sep 9 '15 at 20:22
• @Aretino you said : $7\times8\times3=168$ could u explain that – Educ Sep 9 '15 at 20:26
• I guess 3 positions for '75' then pick two more different numbers. – BruceET Sep 9 '15 at 20:28

If you can't use 7 and 5 you have eight digits left and there are $8\times7$ ways to choose from them the other two digits. As you wrote, there are three possible arrangements for these digits ($75xy$, $x75y$ and $xy75$) so the total number of codes is $56\times3$.