The numbers in the 5th row are" 1,4,6,4,1". If I want to know the numbers in the 72nd row, how can i get the numbers?
The 72nd row should be: 1, 71, ..... ...... ,71, 1. The odd-number (2n-1) row should be: 1, (2n-2), .... ...., (2n-2),1 The even-number (2n) row should be: 1, (2n-1), .... .... , (2n-1), 1
The Pascal's Triangle
the 1st row: 1
the 2nd row: 1, 1
the 3rd row: 1, 2,1
the 4th row: 1,3,3,1
the 5th row: 1, 4,6,4,1
the 6th row: 1,5,10,10,5,1
etc.
As written in the comments, if you're looking for the number in the $n$th row and $k$th column of Pascal's triangle (note that the first row is $n=0$ and the first column is $k=0$), you can use the formula
$${n \choose k}=\frac{n!}{k!(n-k)!}$$
For example, we know that the first column of the $72$nd row will have a $1$, but we can prove this because
$${72 \choose 0}=\frac{72!}{0!(72-0)!}=1$$
Continuing on, we have
$${72 \choose 1}=\frac{72!}{1!(72-1)!}=72$$
$${72 \choose 2}=\frac{72!}{2!(72-2)!}=\frac{72 \times 71 \times 70!}{2 \times 70!}=2556$$
and so on, until $k=72$.
NOTE: If you'd like to know what ${n \choose k}$ represents, it's the number of ways you can select $k$ items from $n$ items, with order not mattering. Each way is called a combination. For example, there are $5$ vowels in the English alphabet, and if I want to know how many combinations of $3$ vowels are possible, I compute ${5 \choose 3}=\frac{5!}{3! \times 2!} = 10$ possible combinations. When I say "order not mattering," in this case that means that, for example, A E O and O A E are considered the same.