Given that \(\binom{15}{8}=6435\), \(\binom{16}{9}=11440\), and \(\binom{16}{10}=8008\), find \(\binom{15}{10}\).

I know that I should be able to do this but I can't figure it out.

Guest Aug 10, 2019

#1**+3 **

**Given that \(\dbinom{15}{8}=6435\), \(\dbinom{16}{9}=11440\) , and \(\dbinom{16}{10}=8008\), find \(\dbinom{15}{10}\).**

**Formula: \(\dbinom{n}{k} + \dbinom{n}{k+1} = \dbinom{n+1}{k+1}\)**

\(\begin{array}{|lrcll|} \hline & \mathbf{\dbinom{n}{k} + \dbinom{n}{k+1}} &=& \mathbf{\dbinom{n+1}{k+1}} \\\\ (1) & \dbinom{15}{8} + \dbinom{15}{9} &=& \dbinom{16}{9} \\ (2) & \dbinom{15}{9} + \dbinom{15}{10} &=& \dbinom{16}{10} \\ \hline (1)-(2): & \dbinom{15}{8} + \dbinom{15}{9} -\left( \dbinom{15}{9} + \dbinom{15}{10}\right) &=& \dbinom{16}{9}-\dbinom{16}{10} \\ & \dbinom{15}{8} - \dbinom{15}{10} &=& \dbinom{16}{9}-\dbinom{16}{10} \\ & 6435 - \dbinom{15}{10} &=& 11440-11440 \\ & 6435 - \dbinom{15}{10} &=& 3432 \\ & \dbinom{15}{10} &=& 6435 - 3432 \\ & \mathbf{\dbinom{15}{10}} &=& \mathbf{3003} \\ \hline \end{array}\)

heureka Aug 11, 2019

#2

#4**0 **

You can limit the number of formulas that you need to memorize by knowing and understanding where the formulas come from.

There are still a lot that it is better just to memorize of course but I know a lot less formulas than most other people of my level.

Most times this does not go against me.

For instance, the distance and midpoint formulas look aweful but if you understand where they come from you do not need to know the formulas at all!

The midpoint formula is just (the average of xs, the average of ys) which is totally logical.

And the distance formula is just pythagoras's theorem.

Melody
Aug 12, 2019