You need to remember how to generate each row of Pascal's triangle using binomial expansion such that:

`n=1 =gt (a+b)^1 = C_1^0*a+C_1^1*b = 1*a+1*b`

You need to substitute 1 for a and b such that:

`(1+1)^1 = C_1^0*1+C_1^1*1 = 1+1 = 2`

The single place in the first row of Pascal triangle is taken by the term 1.

The second row has two terms and they are 1 and 1. Using combinations or binomial coefficients you should substitute `C_1^0` and `C_1^1 ` for these terms 1 and 1.

The third row has three terms such that: `1 2 1` .

Notice that you may find the middle term using the terms in the second row such that: `1+1 = 2` .

Using combinations or binomial coefficients you should substitute `C_2^0` and `C_2^2` for the end terms 1 and 1 and `C_2^1` for the middle term 2.

The fourth row has four terms such that: 1 3 3 1.

Using combinations or binomial coefficients you should substitute `C_3^0 ` and `C_3^3 ` for the end terms 1 and 1 and `C_3^1` and `C_3^2` for the middle terms 3 and 3.

The fifth row has five terms such that: `146 41` .

Using combinations or binomial coefficients you should substitute `C_4^0` and `C_4^4 ` for the end terms 1 and 1 and `C_4^1,C_4^2,C_4^3` for inner terms `4,6,4` .

6th row - `1 5 10 10 5 1`

7th row -`1 6 15 20 15 6 1`

8th row - `1 7 21 35 35 21 7 1`

9th row - `1 8 28 56 70 56 28 8 1`

10th row - `1 9 36 84 126 126 84 36 9 1`

**Hence, the tenth row has 10 terms and using combinations or binomial coefficients you should substitute `C_9^0` and `C_9^9` for the end terms 1 and 1 and `C_9^1,` `C_9^2,C_9^3,C_9^4,C_9^5,C_9^6,C_9^7,C_9^8` for inner terms `9, 36, 84, 126, 126, 84, 36, 9` .**