Using the Pascal's Triangle!

Question

12. Find the 8th number in the 11th row (n=11) of Pascal's triangle.. My book isn't very helpful when explaining this..

Answer 1

                 

   0.                       1

   1                      1  1

   2.                   1  2  1

   3.                 1  3   3   1

   4                 1 4   6   4   1

   5               1  5 10  10  5   1

   6             1  6 15 20   15   6  1

   7            1  7 21  35  35.  21   7  1

   8          1 8 28   56  70  56   28 8  1

   9          1  9 36  84  126 126  84 36 9  1

  10      1 10 45 120 210 252 210 120 45 10 1

  11    1 11 55 165 330 462 462 330 165 55 11 1

 

The triangle builds as shown above and explained 

in previous answers.  The numbers correspond to

the coefficients of the expansion of (a+b)ⁿ 

Each row befins and ends with 1, with subsequent 

numbers being the sum of every pair in the preceding line.

 

Your problem with the statement is that 

Pascals Triangle starts from (a+b)⁰ which

corresponds to n=0.   Thus, n=11 is actually

row 12.  

 

The 8th number in the 11th row is 120

The 8th number corresponding to n=11 is 330

 

You tell me which you meant.

 

Answer 2

I will show you the pattern of Paschal's triangle, and you can take it from there.  Each row starts and ends with a 1, and each number is the sum of the nearest numbers in the row above.  The first row has only a 1.  The second row has a 1 and a 1.  The third row has 3 numbers: 1, 1+1 = 2, 1.  The 4th row has 1, 1+2 = 3, 2+1 =3, 1.  Can you see the pattern?  Below I show you the first 6 rows of the pattern.  Note the symmetry of the triangle.  Can you take it from there up to row 11?

 

       1

      1 1

     1 2 1

    1 3 3 1

  1 4 6 4 1

1 5 10 10 5 1

Answer 3

Each number in the n=11 (typo fixed...was not eleven!) row is computed by 11Cr where r runs from 0 up through 11.

 

The 8th number, counting from the left, is associated with r=7.

 

Now can you do it, using the formula for nCr?