Question

How many entries of Pascal's triangle are equal to 91?

Answer 1

I'll give you a hint. 41C1=41. What are the other possibilities?

Answer 2

Hmm, I dunno how to find other combinations equal to 41.

Answer 3

Unless you are asking me to find something else

Answer 4

There's one other. Imagine you're looking at the triangle. You know it's symmetric. So what's the combination of the opposite side of 41C1?

Answer 5

41c40?

Answer 6

bingo.

Answer 7

So two?

Answer 8

I believe those should be all. It would be little trickier if 41 weren't prime, but since it is, there should only be 2.

Answer 9

Also, what does it mean by "entries"

Answer 10

Entries would just be the different numbers that appear.

Answer 11

Alright, thanks

Answer 12

You're welcome.

Answer 13

What if we are looking for 91, instead of 41?

Answer 14

Because 91 isn't prime

Answer 15

Well \(91=7\cdot13\). So we might get a few subtleties. Give me a minute to check.

Answer 16

Well, if nCr=91=7*13, then we know that n=7,13,91. If it's 91, there are two options.

If n=7, then we won't ever get a 13 in the numerator, so there are no possibilities there.

If n=13, there's only one factor of 7 in \(13!\) (there would be more if we had \(14!\)). And no matter what we divide by, there's always a 7 in the denominator. so there are no possibilities here either.

Answer 17

If that's a bit confusing, you can easily just try every possibility for n=7,13 to verify that there are no other possibilities.

Answer 18

Okay, I think I understand that part.

Answer 19

Right. That comes from when \(n=91\). Then we get the same two possibilities as before. \(c=1,90\).

Answer 20

I thought I described the solution above?

Answer 21

I don't understand the second part

Answer 22

Can someone explain the second part of KingGeorge's solution?

Answer 23

Right. That comes from when n=91. Then we get the same two possibilities as before. c=1,90.
This part I am confused about.

Answer 24

What that was saying (perhaps a bit unclearly) was that 91C1 and 91C90 both equal 91. So just like 41, there are only two entries that are 91.

Answer 25

It says that it is incorrect?

Answer 26

hmm. Let me think about this.

Answer 27

Ah. I see where my error is. I assumed that n must be either 7,13, or 91. This is not true. In fact, a counterexample is 14C2=91.

Answer 28

So there is also 14c2 and 14c12, but how do we know there aren't more?

Answer 29

We know that if \(n\le13\), then there are no solutions. And there can only be those two solutions for \(n=14\). But, now we can consider the way that the triangle is constructed. We know that each entry is made by adding the two entries above. So nCr=(n-1)Cr+(n-1)C(r-1).

So if we have nC2, for some \(n>14\), then we know that it's greater than 91. Give me minute, and I'll try to draw out a picture to make this a bit clearer.

Answer 30

So the coloring was a little less than I hoped, but this should get the point across. Those two points are from 14C2 and 14C12. The shaded area, is everything in the triangle that must be greater than 14C2 or 14C12.