Odd Triangle

Each such number can be expressed in the form $\dbinom{m}{n}$ , and if $m\neq 2n$ , then $\dbinom{m}{n}=\dbinom{m}{m-n}$ so it appears in a pair. Pairs can be ignored, so the only numbers that occur an odd number of times are $\dbinom{10}{5}$ , $\dbinom{8}{4}$ , $\dbinom{6}{3}$ , $\dbinom{4}{2}$ , $\dbinom{2}{1}$ , $\dbinom{0}{0}$ . These end up all being distinct, so the answer is $252+70+20+6+2+1=\boxed{351}$

This is a fantastic little problem.

Lemma 1

We note that any number that is not of the form: $\binom{2n}{n}$ Appears an even number of times in its row.

Proof

This is simple because, say we have some integers $m\ge n\ge 0$ , then: $\binom{m}{n}=\binom{m}{m-n}$ (The number of ways we can choose $n$ items from a set is the same as the number of ways we can choose $m-n$ items which we exclude.)

Now, the only time that we don't have two counts is when $m=2n$ , since: $m-n=2n-n=n$ QED.

Lemma 2

Now, say the number of form $\binom{2n}{n}$ appears more than once in the triangle, then, it must appear an odd number of times, in total.

Proof

Since, it is of the form $\binom{2n}{n}$ , we must also have: $\binom{2n}{n}=\binom{m}{k}$ For some $m,k$ . Then, by the above lemma, since $m\ne 2k$ , the number appears an even number of additional times, say $2l$ . Clearly $2l+1$ is odd, and we are done.

QED.

So, we have our total, by the above lemmas, to be: $\sum_{n=0}^{5}\binom{2n}{n}=351$

A very pretty solution, indeed.

I am not sure what should be the correct approach but writing down a few rows might help.

$\begin{array}{ c c c c c c c} & & & & & & 1 & & & & & & \\ & & & & & 1 & & 1 \\ & & & & 1 & & 2 & & 1 & & & & \\ & & & 1 & & 3 & & 3 & & 1 \\ & & 1 & & 4 & & 6 & & 4 & & 1\\ & 1 & & 5 & & 10 & & 10 & & 5 & & 1\\ 1 & & 6 & & 15 & & 20 & & 15 & & 6 & & 1\\ \end{array}$ The numbers in the subsequent rows will be greater than 11 (except the first two and last two terms as they are 1 and the row number).

Lets have a look at the numbers in the triangle above. Its clear that 6, 1, 2 and 20 appear a odd number of times or we can also assert that the middle term of an even numbered row appears an odd number of times.

Numbers in the subsequent odd rows will appear twice and hence they won't be counted in our final answer. (1 too appears twice but it is counted in our final answer because overall, it appears an odd number of times.)

We have written down the numbers till 6th row. The only even numbered rows left are 8 and 10. The middle term of these rows are $\binom{8}{5}=70$ and $\binom{10}{6}=252$ .

Adding up the numbers, $1+2+6+20+70+252=\fbox{351}$ .

Take note that according to the binomial theorem, the numbers that appear in the Pascal's triangle are actually equivalent to the coefficients that appear in the binomial theorem.

For example: $(x + y)^2 = x^2 + 2ab + y^2$ , where 1, 2, 1 are the numbers in row 2 of the triangle.

Besides, it is noticeable that in row 2, the number 2 appears only once. Same goes to the number 6 in row 4, and 20 in row 6. The rule is that in every even row, (e.g. row 0, 2, 4, etc) there exist a number that occurs for an odd number of times in the triangle, and the number only appears in the row exactly once. It is located at the center of the even row. Even though it might appear in other rows, it appears in pairs. Hence, the number appears for an odd number of times in the triangle.

To further illustrate this, just look at row 4 and row 6:

Row 4: 1 4 6 4 1

Row 6: 1 6 15 20 15 6 1

The numbers 6 and 20 appears in row 4 and row 6 respectively for exactly once, and the numbers are at the center of row 4 and row 6. Although the number 6 also appears in row 6, it appears in a pair.

We can express these numbers as ${n \choose n/2}$ , where n is the row number.

The sum of numbers are:

${0 \choose 0} + {2 \choose 1} + {4 \choose 2} + {6 \choose 3} + {8 \choose 4} + {10 \choose 5}$

$= 1+ 2+ 6 + 20 + 70 + 252$

$= \boxed{351}$

The the $(k-1)$ st number in the $n$ th row equals $\binom{n}{k}$ . Also, for non-negative integers $n, k$ , $k \leq n$ ,

$\begin{aligned} \binom{n}{k} = \binom{n}{n-k} \end{aligned}$

This means, if $n-k \neq k$ , $\binom{n}{k}$ occurs twice in the same row. But, for even $n$ , when $k = \frac{n}{2}$ , $\binom{n}{k} = \binom{n}{n-k}$

Therefore, except for $\binom{0}{0}, \binom{2}{1}, \binom{4}{2}, \binom{6}{3}, \binom{8}{4}, \binom{10}{5}$ , all numbers occur twice in the Pascal's Triangle.

Summing up the numbers above gives us our answer $\boxed{351}$ .

Simply look at a Pascal's triangle. If you look closely, you will see that the numbers that occur an odd number of times from rows 0 to 11 are the numbers that are in line (downwards) with the first "1" at the topmost. Those numbers are 1, 2, 6, 20, 70 and 252. Taking its sum, 1+2+6+20+70+252= 351. :D

In pascal's triangle the middle row alternatively has numbers that exist odd no. of times.

for up till row 11,they are 1+2+6+20+70+252 = 351

Pascal's Triangle is symmetric about a vertical line cutting through the middle of the triangle. This means that if a number $N$ is not on that line, it appears an even number of times, since for every time it appears on the left, it has a corresponding appearance on the right. Therefore, the only $N$ appearing an odd number of times are those exactly on the middle line - that is, those in the form $\dbinom{2k}{k}$

for some nonnegative integer $k.$ [This is because the number in the $r$ th row and $p$ th position from the left can be written as $\binom{r}{p}.$ ] Since we are considering only rows $0$ through $11,$ we can only take $k = 0, 1, 2, 3, 4, 5.$ Doing so gives us

$N = 1, 2, 6, 20, 70, 252$

which sum to $\boxed{351}.$

as we knew pascal triangle has a pattern like this ,the odd-numbered pascal triangle(s) always have a "middle number" ,that is the number that only appears one time . so if it is listed, it should be like this.

third pascal triangle 's middle number : 2

fifth pascal triangle 's middle number : 6

seventh pascal triangle 's middle number : 20

ninth pascal triangle 's middle number : 70

eleventh pascal triangle 's middle number : 252

and, the number "1" , appears an odd number of times, because, every pascal triangles has two 1's , thus we obtain the number of 1 appeared in the 0-11th pascal triangle

first-eleventh pascal triangles : 2x11 = 22 1's

and the 0th pascal triangle has exactly one 1's ,so there are 23 1's

then we know that 1 is counted

so now we sum the "middle number"(s) we've found : 2+6+20+70+252= 350

and because 1 is counted the sum is : 350+1 = 351

The main problem here is, in my opinion understanding the question. It would have been lot more easier if the question had better punctuation. like

What is the sum of all numbers that occur, an odd number of times, in rows 0 through 11 of Pascal's triangle?

once you have understood the question it is only to check pascals triangle manually. you will find it easier to check pascals triangle row-by-row, not column. thanks