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Difference between revisions of "Pascal Triangle Related Problems"

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m (See Also Section)
 
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=== Problem 1 ===
 
=== Problem 1 ===
  
−
Given <math>\displaystyle (x + y)^{10}</math>, find:
+
Given <math>(x + y)^{10}</math>, find:
  
 
# The coefficient of the <math>x^{4}y^{6}</math> term.
 
# The coefficient of the <math>x^{4}y^{6}</math> term.
Line 34: Line 34:
 
  1  10  45  120  210  252  210  120  45  10  1
 
  1  10  45  120  210  252  210  120  45  10  1
  
−
Thus the coefficient is the 6th number in the row or <math>\displaystyle{210}</math>.
+
Thus the coefficient is the 6th number in the row or <math>{210}</math>.
  
 
This can also be found using the binomial theorem:
 
This can also be found using the binomial theorem:
  
−
::<math>(x+y)^{k}=\sum^{k}_{n=0}\displaystyle{k\choose n}x^{n}y^{k-n}</math>
+
::<math>(x+y)^{k}=\sum^{k}_{n=0}{k\choose n}x^{n}y^{k-n}</math>
  
−
Through the summation, the binomial theorem will provide you with the coefficient if each term of the result.  In our particular case, we are only looking for the coefficient of the <math>\displaystyle{x^{4}y^{6}}</math> term.
+
Through the summation, the binomial theorem will provide you with the coefficient if each term of the result.  In our particular case, we are only looking for the coefficient of the <math>{x^{4}y^{6}}</math> term.
  
−
Since you are looking for <math>\displaystyle{x^{4}y^{6}}</math> term in <math>\displaystyle{(x+y)^{10}}</math>, then <math>\displaystyle{n=4}</math> and <math>\displaystyle{k=10}</math>.
+
Since you are looking for <math>{x^{4}y^{6}}</math> term in <math>{(x+y)^{10}}</math>, then <math>{n=4}</math> and <math>{k=10}</math>.
  
−
So the coefficient of the <math>\displaystyle{x^{4}y^{6}}</math> term is <math>\displaystyle{10\choose 4}=210</math>.
+
So the coefficient of the <math>{x^{4}y^{6}}</math> term is <math>{10\choose 4}=210</math>.
  
 
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:
 
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:
Line 50: Line 50:
 
::<math>\sum_{n=0}^{k}{k \choose n}=2^k</math>
 
::<math>\sum_{n=0}^{k}{k \choose n}=2^k</math>
  
−
This summation formula simply adds up all the coefficients since <math>\displaystyle{k\choose n}</math> gives us each of the coefficients.  So, the sum is <math>2^{10}=1024</math>.
+
This summation formula simply adds up all the coefficients since <math>{k\choose n}</math> gives us each of the coefficients.  So, the sum is <math>2^{10}=1024</math>.
  
−
For your information, the final polynomial which results from <math>(x+y)^{10}</math> is <math>\displaystyle x^{10} + 10x^9y + 45x^8y^2 + 120x^7y^3 + 210x^6y^4 + 252x^5y^5</math><math> + 210x^4y^6 + 120x^3+y^7 + 45x^2+y^8+10x^1y^9+y^{10}</math>
+
For your information, the final polynomial which results from <math>(x+y)^{10}</math> is <math>x^{10} + 10x^9y + 45x^8y^2 + 120x^7y^3 + 210x^6y^4 + 252x^5y^5</math><math> + 210x^4y^6 + 120x^3+y^7 + 45x^2+y^8+10x^1y^9+y^{10}</math>
  
 
== Intermediate ==
 
== Intermediate ==
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== Olympiad ==
 
== Olympiad ==
 +
 +
==See Also==
 +
*[[Pascal's triangle]]
 +
 
{{stub}}
 
{{stub}}
  
 
[[Category:Introductory Combinatorics Problems]] [[Category:Intermediate Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]] [[Category:Intermediate Combinatorics Problems]]

Latest revision as of 20:18, 3 January 2023

The Triangle

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

Row
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

Introductory

Problem 1

Given $(x + y)^{10}$, 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 number in each row is considered the 0th number) of the 10th row in Pascal's triangle.

The 10th row is: 

1  10  45  120  210  252  210  120  45  10  1

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

This can also be found using the binomial theorem:

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

Through the summation, the binomial theorem will provide you with the coefficient if each term of the result. In our particular case, we are only looking for the coefficient of the ${x^{4}y^{6}}$ term.

Since you are looking for ${x^{4}y^{6}}$ term in ${(x+y)^{10}}$, then ${n=4}$ and ${k=10}$.

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

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:

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

This summation formula simply adds up all the coefficients since ${k\choose n}$ 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}$

Intermediate

Problem 1

In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3: 4: 5$?

Solution

Problem 2

A triangular array of squares has one square in the first row, two in the second, and in general, $k$ squares in the $k$th row for $1 \leq k \leq 11.$ With the exception of the bottom row, each square rests on two squares in the row immediately below (illustrated in given diagram). In each square of the eleventh row, a $0$ or a $1$ is placed. Numbers are then placed into the other squares, with the entry for each square being the sum of the entries in the two squares below it. For how many initial distributions of $0$'s and $1$'s in the bottom row is the number in the top square a multiple of $3$?

2007 AIME II-13.png

Solution

Olympiad

See Also

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