what is the relationship between Pascal triangle and grouping?

Answer 1

The terms in Pascal's triangle is given by the formula:

n! / ( k! (n-k)! )

which happens to be the same formula for a combinarial where k is the sample size and n is the number of objects in the set. For example, if I have 5 different cards, how many ways can I pick 2 cards out of the 5? The answer is

5! / ( 2! (5-2)! )

or 120 / 2 * 6, or 10 different ways. That is 2nd binomial term in Pascal's triangle where n = 5. Pascal's triangle generates binomial numbers which are found in powers of sums, so that with this example given, the expression

(x +1)^5

equals

x^2 + 5 x^4 + 10 x^3 + 10 x^2 + 5 x + 1

and 10 would be the 2nd pascal term

Answer 2

It helps with expanding co-efficients. For instance, take an expression such as (x+1) and multiply it a number of times, you'll notice the co-efficients correspond to the levels on Pascal's triangle:

(x + 1)^0 = 1
(x + 1)^1 = x + 1
(x + 1)^2 = x^2 + 2x + 1
(x + 1)^3 = x^3 + 3x^2 + 3x + 1
(x + 1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1

...

You'll notice that the coefficients on all the terms correspond to Pascal's levels on the triangle.

Best of luck!

Answer 3

you're probable considering the identity: Sum from ok = 0 to ok = n of (n - ok choose ok) = the n + 1st Fibonacci decision. to illustrate, for n = 5, we've: (5 choose 0) + (4 choose a million) + (3 choose 2), or a million + 4 + 3 = 8, and eight is the sixth Fibonacci decision. And for n = 6, we've: (6 choose 0) + (5 choose a million) + (4 choose 2) + (3 choose 3), or a million + 5 + 6 + a million = 13, the seventh Fibonacci decision. Now, reckoning on Pascal's triangle, we see that (x choose y) is the x'th row and y'th column of the triangle, and that permits you to initiate on the x'th row and bypass up one row and one column each and each and every time till we can't go any farther. The sum of those numbers stands out because the Fibonacci numbers. wish this helps!

Answer 4

I have wondered that too... I gave up. I don't get it!!