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Pascal Triangle Related Problems

Revision as of 20:04, 3 May 2007 by Levans (talk | contribs) (→‎The triangle)

The triangle

Here are lines zero through twelve of Pascal's triangle: 

Row 0:                                                 1                                              
Row 1:                                              1     1                                           
Row 2:                                           1     2    1                                         
Row 3:                                        1     3     3    1                                      
Row 4:                                     1     4     6     4    1                                   
Row 5:                                  1     5     10    10    5    1                                
Row 6:                               1     6     15    20   15    6    1                             
Row 7:                            1     7     21    35    35    21    7    1                          
Row 8:                         1     8     28    56    70    56    28    8    1                       
Row 9:                      1     9     36    84    126   126   84    36    9    1                    
Row 10:                  1     10    45    120   210   252   210   120   45    10   1                 
Row 11:               1     11    55   165   330    462   462   330   165   55    11  1               
Row 12:            1     12    66    220   495   792   924   792   495   220   66    12  1

Problem 1

You are given the binomial $x+y$. Raise it to the 10th power.

Find:

1. The coefficient of the $x^{4}y^{6}$ term.

2. The sum of the coefficients.

Solution

1. You need to find the 6th number (remember the first column on each row is considered the 0th number) of the 10th row in Pascal's triangle.

The 10th row is: 

Row 10:                  1     10    45    120   210   252   210   120   45    10   1

Thus the coefficient is the 6th number in the row or $\displaystyle{210}$.

This can also be found using the binomial theorem: 

$(x+y)^{k}=\sum^{k}_{n=0}\displaystyle{k\choose n}x^{n}y^{k-n}$

Since you are looking for $\displaystyle{x^{4}y^{6}}$, then $\displaystyle{n=4}$ and $k-n=6 \Rightarrow k-4=6 \Rightarrow k=10$.

So the coefficient of the $x^{4}y^{6}=\displaystyle{10\choose 4}=210$.

2. Since all the coefficients are found in the 10th row, we simply need to add the numbers in the 10th row together. This can be done by hand since there are relatively few numbers, but we could also use the following formula to sum up the numbers: 

$\sum_{k=0}^{n}{{n \choose k}}=2^n$

This summation formula simply adds up all the coefficients since $\displaystyle{k\choose n}$ gives us each of the coefficients. So, the sum is $2^{10}=1024$.

For your information, the final polynomial which results from $(x+y)^{10}$ is: 

$x^{10} + 10x^9y + 45x^8y^2 + 120x^7y^3 + 210x^6y^4 + 252x^5y^5 + 210x^4y^6 + 120x^3+y^7 + 45x^2+y^8+10x^1y^9+y^{10}$
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