1
$\begingroup$

I was reading a text book and found these two lines and I have no clue how did step1 become step2. Please help me with this. Thanks

Step 1:

$$ \frac {5!}{(4-r)!} = \frac {6*5!}{(5-r+1)(5-r)(5-r-1)!} $$

Step 2: $$ (6-r)(5-r)=6 $$

Text book: (NCERT 11th Chapter 7, Example 13)

$\endgroup$
2
$\begingroup$

First things first: Simplify.

If we simplify the first step, we have $$\frac{5!}{(4-r)!}=\frac{6 \times 5!}{(6-r)(5-r)(4-r)!}$$

Now multiply each side by $\frac{(6-r)(5-4)(4-r)!}{5!}$.

Thus we have $$\frac {5!}{(4-r)!} = \frac {6*5!}{(5-r+1)(5-r)(5-r-1)!} \Leftrightarrow (6-r)(5-r)=6$$

$\endgroup$
  • $\begingroup$ wow, thats wonderful, thanks. I also want to know another thing. whats logic behind rewriting (5-r+1)! into (5-r+1)(5-r)(5-r-1)! $\endgroup$ – pymacstue Mar 28 '16 at 9:04
  • $\begingroup$ @PYmacstue It makes it easier to simplify. The right side has $(4-r)!$, so we have to multiply each side by $(4-r)!$. But if the right side if $(6-r)!$, It's harder to see. $\endgroup$ – S.C.B. Mar 28 '16 at 9:06

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.