Question: Explain why an entry in Pascal's Triangle is the sum of the two entries diagonally above it.

See answer below: ALSO: Chapter 6 is all combinatorics! The very first word in the chapter is "Combinatorics"! (In other words, the question shows a much deeper problem than just not getting question 17: the student is not even aware of a major topic covered during the course, and probably has never looked inside the textbook.)

This was covered in the lectures in ALEC and BLEC.

Consider a scenario where you have 2 switches on a board. The cases for the 2-switch board will be

a) 1 possibility of

Both OFF (00)

b)1 switch ON which has these 2 possibilities

1 ON 1 OFF (01)

1 OFF 1 ON (10)

c)1 possibility of

Both ON (11)

The above cases a, b and c sum up the 3rd row of the Pascal Triangle " 1 2 1".

Now say you want to add one more switch to case b such that again 1 switch is ON in the 3-switch board (you cannot touch the 2 switches from the previously set 2-switch board).

To do so,

you need to add one more switch which is OFF.

The OFF case from above scenario is case a.

Therefore the new combinations available will be 010, 001 or 100 such that one switch is ON.

Number of possible combinations are 3 which is the addition of values 1 and 2 present diagonally above the value 3.

You cannot pick case c for one switch to be ON because case c already has 2 switches ON  and our aim is to have just 1 switch ON.

                                                1

                                1                                1

                     00  1 01        2   10                         1   11

        1                                    3                         3                        1

                          001   010    100 (1 bit ON)

Consider the expansion of (a+b)3

We have multiplication of 3 terms

(a+b) * (a+b) * (a+b)

If we pick up just a each from one term how many possibilities are there?

(a+b) * (a+b) * (a+b)

→ Possibility: 1

Similarly, if we pick b from each term

(a+b) * (a+b) * (a+b)

→ Possibility: 1

If we decide to pick 2 a’s and 1 b each from one term

(a+b) * (a+b) * (a+b)

(a+b) * (a+b) * (a+b)

(a+b) * (a+b) * (a+b)

→ Possibilities: 3

Similarly, if we pick 2 b’s and 1 a from each term

(a+b) + (a+b) + (a+b)

(a+b) + (a+b) + (a+b)

(a+b) + (a+b) + (a+b)

→ Possibilities: 3

The above possibilities give us the coefficients for the binomial expansion of (a+b)3

→ 1a3 + 3a2b + 3ab2 + 1b3