-3
$\begingroup$

The questions asks to prove combinatorially.

${10}\choose{5}$ = ${4}\choose{4}$ + ${5}\choose{4}$ + ${6}\choose{4}$ + ${7}\choose{4}$ + ${8}\choose{4}$ + ${9}\choose{4}$

So I know that the LHS is choosing 5 objects from 10 objects.

If we let there be 4 objects and say 6 special objects.

The right side is:

${4}\choose{4}$ = Choose 4 from a 4 objects.

${5}\choose{4}$ = Choose 4 from 4 objects and 1 Special Object

${6}\choose{4}$ = Choose 4 from 4 objects and 2 special objects

${7}\choose{4}$ = Choose 4 from 4 objects and 3 special objects

${8}\choose{4}$ = Choose 4 from 4 objects and 4 special objects

${9}\choose{4}$ = Choose 4 from 4 objects and 5 special objects

Thus the right side counts the same thing as the left side. Does this work as a combinatorial proof? If not how could I improve my approach. Thank you.

$\endgroup$
  • 1
    $\begingroup$ I have no idea what you mean by special objects, and so far you haven't done anything. You want to chose 5 numbers out of the set 1, 2, \dots, 10. The last number chosen is 10, 9, 8, 7, 6, or 5. Hence... $\endgroup$ – fredgoodman Mar 17 '18 at 20:28
  • $\begingroup$ @fredgoodman Suppose, instead you had to choose 10 children and form a team of 5. These 10 children include 4 boys and 6 girls. Does that clear it up slightly? $\endgroup$ – Safder Aree Mar 17 '18 at 20:31
  • $\begingroup$ It makes it clear that you are quite confused, as you have just told the story which goes with a completely different identity. $\endgroup$ – fredgoodman Mar 17 '18 at 20:45
  • $\begingroup$ Yes. I am confused. $\endgroup$ – Safder Aree Mar 17 '18 at 20:47
  • $\begingroup$ I am sorry you are confused. You may need to talk to someone face to face. Or else quietly ponder the hint given by Foobaz John. $\endgroup$ – fredgoodman Mar 17 '18 at 20:48
4
$\begingroup$

Hint

Partition the $5$ element subsets of $\Omega=\{1,2,\dotsc,10\}$ based on their maximum element. How many $5$ element subsets of $\Omega$ have $10$ as their maximum element, $9$ as their maximum element and so on?

$\endgroup$
  • $\begingroup$ I'm having trouble understanding what to do here. $\endgroup$ – Safder Aree Mar 17 '18 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.