Table of Contents

Last modified on August 3rd, 2023

Pascal’s triangle is a number triangle that starts with 1 on the top and continues such that each row has 1 at its two ends. It is named after Blaise Pascal, a 17th-century famous French mathematician and philosopher.

We consider the left-most element in each row as the 0^{th} element. Then the element next to the right of the 0^{th} element is the 1^{st} element. Similarly, the element next to the 1^{st} element is the 2^{nd} element, and so on.

In Pascal’s triangle, the middle numbers are filled such that it is the sum of those adjacent numbers placed just above it.

We consider the adjacent elements to 1 in every row as 0 because there is nothing next to 1 in each row. We repeat this method to construct each subsequent row.

The same process is repeated to fill up the other elements as the Pascal’s triangle has infinite rows.

Pascal’s triangle has different patterns within the triangle. Some of them are listed:

- The first diagonal is ‘1’
- The next diagonal contains the counting or
**natural numbers**(1, 2, 3,……) - The third diagonal contains the
**triangular numbers**(1, 3, 6, 10, 15,……) - The fourth diagonal contains
**tetrahedral numbers**(1, 4, 10, 20,……) - The fifth diagonal contains
**pentatope numbers**(1, 5, 15, 35,……) and so on.

In Pascal’s triangle, the numbers on the left side are identical to those on the right, just like a mirror image. Thus the triangle is symmetrical.

If we add the numbers in each row, the value doubles each time compared to the previous row. The sums are all powers of 2 and thus can be represented as 2^{k} for the k^{th} row, k = 0, 1, 2, 3,……

We write the summation identity as ${\sum ^{k}_{n=0}\begin{pmatrix} k \\ n \end{pmatrix}}$ = 2^{k}

**Find the sum of all elements in the 4 ^{th} row of Pascal’s Triangle.**

Solution:

As we know,

${\sum ^{k}_{n=0}\begin{pmatrix} k \\ n \end{pmatrix}}$ =2^{k}

Here, the sum of all elements in the 4^{th} row will be,

2^{4 } = ${\sum ^{4}_{n=0}\begin{pmatrix} 4 \\ n \end{pmatrix}}$ = ${\begin{pmatrix} 4 \\ 0 \end{pmatrix}}$ + ${\begin{pmatrix} 4 \\ 1 \end{pmatrix}}$ + ${\begin{pmatrix} 4 \\ 2 \end{pmatrix}}$ + ${\begin{pmatrix} 4 \\ 3 \end{pmatrix}}$ + ${\begin{pmatrix} 4 \\ 4 \end{pmatrix}}$

= 1+ 4 + 6 + 4 + 1 = 16

**Work out the sum of all elements in the 10 ^{th} row of Pascal’s Triangle.**

Solution:

As we know,

${\sum ^{k}_{n=0}\begin{pmatrix} k \\ n \end{pmatrix}}$ = 2^{k}

Here, the sum of all elements in the 10^{th} row will be,

2^{10} = ${\sum ^{10}_{n=0}\begin{pmatrix} 10 \\ n \end{pmatrix}}$ = ${\begin{pmatrix} 10 \\ 0 \end{pmatrix}}$ + ${\begin{pmatrix} 10 \\ 1 \end{pmatrix}}$ + ${\begin{pmatrix} 10 \\ 2 \end{pmatrix}}$ + ${\begin{pmatrix} 10 \\ 3 \end{pmatrix}}$ + ${\begin{pmatrix} 10 \\ 4 \end{pmatrix}}$ + ${\begin{pmatrix} 10 \\ 5 \end{pmatrix}}$ + ${\begin{pmatrix} 10 \\ 6 \end{pmatrix}}$ + ${\begin{pmatrix} 10 \\ 7 \end{pmatrix}}$ + ${\begin{pmatrix} 10 \\ 8 \end{pmatrix}}$ + ${\begin{pmatrix} 10 \\ 9 \end{pmatrix}}$ + ${\begin{pmatrix} 10 \\ 10 \end{pmatrix}}$

= 1 + 10 + 45 + 120 + 210 + 252 + 210 + 120 + 45 + 10 + 1 = 1024

Each line of Pascal’s triangle is the powers or the exponents of 11.

- 11⁰ = 1 (the first row contains only 1)
- 11¹ = 11 (the second row has 1 and 1)
- 11² = 121 (the third row has 1, 2, 1) and so on.

But for 11⁵, how can we relate to the above pattern?

Here, the digits overlap, like shown below.

The same thing happens with 11⁶ too.

The rest of the triangle follows the same pattern.

In the second diagonal of Pascal’s triangle, the square of a number is equal to the sum of the numbers next to it and below both of them.

For example,

3 (from third diagonal) + 6 (from third diagonal) = 9 = 3^{2 }(3 from second diagonal)

We can obtain a Fibonacci series in Pascal’s Triangle by summing the different diagonal elements of the triangle as shown.

Suppose we shade the even and odd elements of Pascal’s triangle (marked in different colors). In that case, we will get a pattern similar to the Sierpinski Triangle.

In Pascal’s Triangle, each entry represents the number of paths from the top down.

For example, there is only one path from the top down to any 1, two different paths to 2, and 3 pathways to 3.

Another exciting pattern visible in the triangle deals with prime numbers. Suppose a row begins with a prime number or is a prime-numbered row. Then, all the numbers in that row except the 1’s are divisible by that prime number. Looking at the 7^{th} row (1 7 21 35 35 21 7 1), we can see that 7, 21, and 35 all are divisible by 7.

However, this property does not hold true for the composite numbers. For example, if we look at row 8 (1 8 28 56 70 56 28 8 1), we can see that 28 and 70 are not divisible by 8 (see example 2 diagram).

To find the value of any element, say the j^{th} element of k^{th} row of Pascal’s triangle, we use the formula:

x_{k,j}= ${\begin{pmatrix} k \\ j \end{pmatrix}}$

Here ${\begin{pmatrix} k \\ j \end{pmatrix}}$ =${\begin{pmatrix} k-1 \\ j-1 \end{pmatrix}}$ +${\begin{pmatrix} k-1 \\ j \end{pmatrix}}$ and j is any non negative integer such that 0${\leq}$j${\leq}$k

The above formula is also written as:

${\begin{pmatrix} k \\ j \end{pmatrix}}$ (i.e. , k choose j) = C(k,j) =^{k}C_{j} = ${\dfrac{k!}{j!\left( k-j\right) !}}$

Here =^{ k}C_{j} denotes the j^{th} element in the k^{th} row.

The elements of Pascal’s Triangle is also be written as: (illustration is given up to row 7, starting from row 0)

**Find the 3rd element of the 5 ^{th }row in Pascal’s triangle.**

Solution:

As we know,^{k}C_{j} = ${\begin{pmatrix} k \\ j \end{pmatrix}}$ =${\begin{pmatrix} k-1 \\ j-1 \end{pmatrix}}$ + ${\begin{pmatrix} k-1 \\ j \end{pmatrix}}$ = ^{k-1}C_{j-1} + ^{k-1}C_{j}

Here k = 5 and j = 3 so, we have to calculate ^{5}C_{3}

Using the above formula we get ^{5}C_{3} = ^{(5-1)}C_{(3-1)} + ^{(5-1)}C_{3}=^{4}C_{2}+ ^{4}C_{3}= 2^{nd}element of 4^{th} row + 3^{rd} element of 4^{th} row =6 + 4 =10 .

In Pascal’s Triangle, each number represents the coefficient of the terms of binomial expansion (x+y)^{n}, where x and y are any two variables and n = 0,1,2,……..

Now expanding (x+y)^{n}, we get,

(x+y)^{n }= a_{0}x^{n }+ a_{1}x^{(n-1)}y + a_{2}x^{(n-2)}y^{2}+………..+a_{(n-1)}xy^{(n-1) }+ a_{n}y^{n}

Here, the coefficients of the form a_{k}, k = 0,1,2,…..n are the elements in the n^{th} row of Pascal’s Triangle and can be expressed as a_{k} =${\begin{pmatrix} n \\ k \end{pmatrix}}$ =^{n}C_{k} =^{(n-1)}C_{(k-1) }+ ^{(n-1)}C_{k}

For example, if we expand the expression (x+y)^{n} for n=4,

(x+y)^{4} =a_{0}x^{4} + a_{1}x^{3}y + a_{2}x^{2}y^{2} + a_{3}xy^{3} + a_{4}y^{4}

= ^{4}C_{0}x^{4} + ^{4}C_{1}x^{3}y +^{4}C_{2}x^{2}y^{2} +^{4}c_{3}xy^{3} + ^{4}C_{4}y^{4}

=(1) x^{4} + (4)x^{3}y + (6)x^{2}y^{2} + (4)xy^{3} + (1)y^{4}

Here the coefficients 1, 4, 6, 4, 1 represent elements in the 4^{th} row.

**Find all the coefficients of the expansion of the polynomial (x+y) ^{7} using Pascal’s triangle**.

Solution:

The coefficients of the expansion of the (x+y)^{7 }will actually be the elements in row 7 of Pascal’s triangle.

Elements in the 7^{th} row of Pascal’s Triangle are 1, 7, 21, 35, 35, 21, 7,1 .

So the coefficients of the expansion of (x+y)^{7 }are 1, 7, 21, 35, 35, 21, 7,1

Pascal’s triangle has several real-world applications, such as in mathematical calculations, including probability, combinations, etc.

Pascal’s triangle is also used to find various probability conditions. For example, if we toss a coin just once, there are only two possible outcomes, either Head (H) or Tail (T).

If we toss it two times, there is one possibility of getting both heads HH or tails TT, but there are two possibilities of getting at least a head or tail, i.e., HT or TH.

Now if we consider Pascal’s Triangle, we can see the table based on the number of tosses and outcomes given below:

Number of times a coin is tossed (Row of Pascal’s Triangle) | Outcomes in Combinations | Elements in Pascal’s Triangle |
---|---|---|

1 | {H}, {T} | 1, 1 |

2 | {HH}, {HT, TH}, {TT} | 1, 2, 1 |

3 | {HHH}, {HHT, HTH, THH} {HTT, THT, TTH} {TTT} | 1, 3, 3, 1 |

4 | {HHHH}, {HHHT, HHTH, HTHH, THHH} {HHTT, HTHT, HTTH, THTH, THHT, TTHH} {HTTT, THTT, TTHT, TTTH} {TTTT} | 1, 4, 6, 4, 1 |

…etc… | ….etc…. | …..etc….. |

**Find the probability of getting exactly two heads when a coin is tossed four times.**

Solution:

As we know,

${\sum ^{k}_{n=0}\begin{pmatrix} k \\ n \end{pmatrix}}$ = 2^{k}, for k^{th} row, k = 0, 1, 2, 3,……

Now we know that the number of times a coin is tossed = Row of Pascal’s Triangle

So, the number of total outcomes = 2^{4 }= 16 = (1 + 4 + 6 + 4 + 1), where 6 out of them give exactly two heads. So, the probability of getting exactly two heads is 6/16 or 37.5%.

Pascal’s triangle also shows the different ways by which we can combine its various elements.

The number of ways r number of objects is chosen out of n objects irrespective of any order and repetition is given by:

^{n}C_{r}= ${\dfrac{n!}{r!\left( n-r\right) !}}$, which is the r^{th} element of the n^{th} row of Pascal’s Triangle.

Suppose we have 15 pool balls. How many ways can we choose just 5 of them (ignoring the order in which we select them)?

Here, n = 15, r = 5

If we go down to the start of row 15 and then along 5 places to the right, we find the answer 3003, which is the same if we calculate ^{15}C_{5} separately.

In Pascal’s triangle, let us start at any of the ‘1’ elements (apart from the topmost row) and then move on the left or right side along the row.

Suppose we sum the elements diagonally in a straight line and stop anywhere. In that case, we will find that the next element down diagonally in the opposite direction will equal the sum. When we circle these elements in the triangle, these create a shape like a ‘hockey stick’.

If we begin at the p^{th} row and end on the q^{th} row, the sum will be

${\sum ^{q}_{k=p}\begin{pmatrix} k \\ p \end{pmatrix}}$ = ${\begin{pmatrix} q+1 \\ p+1 \end{pmatrix}}$

**Find ${\sum ^{6}_{k=3}\begin{pmatrix} k \\ 3 \end{pmatrix}}$ for Pascal’s Triangle.**

Solution:

As we know, that by Hockey Stick identity

${\sum ^{q}_{k=p}\begin{pmatrix} k \\ p \end{pmatrix}}$ = ${\begin{pmatrix} q+1 \\ p+1 \end{pmatrix}}$, here p = 3 and q = 6

Here, we begin at the 3^{rd} row and end at the 6^{th} row,

${\sum ^{6}_{k=3}\begin{pmatrix} k \\ 3 \end{pmatrix}}$ = ${\begin{pmatrix} 6+1 \\ 3+1 \end{pmatrix}}$ =${\begin{pmatrix} 7 \\ 4 \end{pmatrix}}$ =35

Also, let’s look at the diagram, summing the elements starting from 1 on the 3^{rd} row and going diagonally down in a straight line up to the 6^{th} row. We can see 1 + 4 + 10 + 20 = 35, the next element down diagonally in the opposite direction.

Last modified on August 3rd, 2023