# An introduction to AND gates - Answers

**Q1.** When we have just 2 inputs, there are a total of 4 combinations of 1s and 0s for the inputs (2 to the power 2). If you had 3 inputs, there are 8 possible combinations. (2 to the power 3).**Q2.** If you had 4 inputs, there would be 16 possible combinations (2 to the power 4).**Q3.** The truth table for this diagram ....

A |
B |
D |
C |
E |

0 | 0 | 0 | 0 | 0 |

0 | 0 | 1 | 0 | 0 |

0 | 1 | 0 | 0 | 0 |

0 | 1 | 1 | 0 | 0 |

1 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 0 | 0 |

1 | 1 | 0 | 1 | 0 |

1 | 1 | 1 | 1 | 1 |

Something very interesting is happening in columns A, B and D. We are counting in binary to get the different permutations.

000 - 0

001 - 1

010 - 2

011 - 3

100 - 4

101 - 5

110 - 6

111 - 7

**Q4.** There are 4 inputs in this diagram, A, B, D and F.

**Q5.** There 2 to the power of 4 = 16 different combinations of inputs.**Q6.** There is only one output to the whole diagram, G. There are outputs from other logic gates though. These are C and E. **Q7.** The truth table is as follows:

A |
B |
D |
F |
C |
E |
G |

0 | 0 | 0 | 0 | 0 | 0 | 0 |

0 | 0 | 0 | 1 | 0 | 0 | 0 |

0 | 0 | 1 | 0 | 0 | 0 | 0 |

0 | 0 | 1 | 1 | 0 | 0 | 0 |

0 | 1 | 0 | 0 | 0 | 0 | 0 |

0 | 1 | 0 | 1 | 0 | 0 | 0 |

0 | 1 | 1 | 0 | 0 | 0 | 0 |

0 | 1 | 1 | 1 | 0 | 0 | 0 |

1 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 0 | 1 | 0 | 0 | 0 |

1 | 0 | 1 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 1 | 0 | 0 | 0 |

1 | 1 | 0 | 0 | 1 | 0 | 0 |

1 | 1 | 0 | 1 | 1 | 0 | 0 |

1 | 1 | 1 | 0 | 1 | 1 | 0 |

1 | 1 | 1 | 1 | 1 | 1 | 1 |

We are counting in binary in columns A, B, D and F, again! This will give us all of the permutations we need to try out!

0000 - 0

0001 - 1

0010 - 2

0011 - 3

0100 - 4

0101 - 5

0110 - 6

0111 - 7

1000 - 8

1001 - 9

1010 - 10

1011 - 11

1100 - 12

1101 - 13

1110 - 14

1111 - 15

**Q8.** This diagram also has 4 inputs.

**Q9.** That means there are 2 to the power of 4 = 16 different permutations of inputs.**Q10.** Here is the truth table. N0te that you could layout the 1s and 0s in columns D and E in the same way as A and B, but if you understand counting in binary, it makes getting all the permutations down in the table quick and easy.

A |
B |
D |
E |
C |
F |
G |

0 | 0 | 0 | 0 | 0 | 0 | 0 |

0 | 0 | 0 | 1 | 0 | 0 | 0 |

0 | 0 | 1 | 0 | 0 | 0 | 0 |

0 | 0 | 1 | 1 | 0 | 1 | 0 |

0 | 1 | 0 | 0 | 0 | 0 | 0 |

0 | 1 | 0 | 1 | 0 | 0 | 0 |

0 | 1 | 1 | 0 | 0 | 0 | 0 |

0 | 1 | 1 | 1 | 0 | 1 | 0 |

1 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 0 | 1 | 0 | 0 | 0 |

1 | 0 | 1 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 1 | 0 | 1 | 0 |

1 | 1 | 0 | 0 | 1 | 0 | 0 |

1 | 1 | 0 | 1 | 1 | 0 | 0 |

1 | 1 | 1 | 0 | 1 | 0 | 0 |

1 | 1 | 1 | 1 | 1 | 1 | 1 |

It should be clear to students now that all the inputs must be a 1 if the output is to be a one, in these examples using just AND gates. It's good practice to get them to draw out the truth tables and start thinking in binary as well as logic.