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)


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

  • $\begingroup$ 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

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