Question

Find the value of r with steps.

7Pr−7Cr=175 [Permutations and combinations]



to check your answer r = { 3 }

Another question down/

Answer 1

\[7\Pr - 7Cr = 175\]

Answer 2

Is r a subscript?

Answer 3

Yep.

Answer 4

I'm not really sure what P and C are. Are these like combinatorics or is this like a recurrence relation? Haha sorry to ask so many questions, I just see letters and don't really know what you're asking but I saw the android so I was interested in helping out a programmer. =P

Answer 5

No problem. They are Permutations and combinations.

Answer 6

Well how far are you able to get on your own? Maybe the definitions will help: \[\Large _nP_r = \frac{n!}{(n-r)!}\] \[\Large _nC_r = \frac{n!}{(n-r)! * r!}\]

Answer 7

This isn't the problem. Whatever I do. Nothing changes.

Answer 8

I reach a stalemate

Answer 9

So you're looking for a strategy of how to solve this?

Answer 10

I do.

Answer 11

Hmm, well honestly I can't say that I have an exact method for figuring this particular example out other than plugging in the values, factoring out a 7! and dividing both sides giving us 5/144 on the right side. \[\Large \frac{1}{(7-r)!} \left( 1- \frac{1}{r!} \right)= \frac{5}{144} \] So reararrange a little to get \[\Large \frac{1}{(7-r)!} \left( \frac{r!-1}{r!} \right)= \frac{5}{144} \] Since the top is 5, I can just isolate this\[\Large r!-1=5\]\[\Large r!=6\] Which from experience I know is r=3. You have to sort of be able to reason out the fact that there are integers on top and bottom and you don't have a fraction as the result of a factorial, and that's how you're able to separate it out. There may be a better way but this is how I figured it out.

Answer 12

Sorry but I have a small objection 5/144 may have been 10/288 or other number
so to take the top with the top you have to add constant (5n/144n) which is not equal to zero.
r! -1 = 5n ( n is a constant \[\neq 0\])
am i mistaken ?

Answer 13

Well that might be true but the fact that we have (7-r)! in the expression indicates that r has to be somewhere from r=0 to r =7 which is a small range to check. I guess I got lucky though.

Even if we formulate like you say, which I have to admit is a really good catch, we would have: \[\Large r!=5n+1\] And now we can easily try to check on our domain of values what possible ways we can multiply consecutive numbers and it be represented as 5n+1, which narrows it down further to n=0, n=1 and those are both pretty easy to check I guess.

Answer 14

Well, so this question don't have any direct answer right ?

Answer 15

It has only one single answer, but it is not like maybe some other algebra problems you've seen where you could blindly follow rules to get there. You will have to think and reason to get to this answer, definitely.

Answer 16

Thanks . Could you help me with another math problem ( easier one).

Answer 17

Yeah sure haha, why not.

Answer 18

Find the limit if they exist:
f(x) = \[\frac{ asin(x) - xsin(a) }{ x - a } x < a\]
f(x) = \[\frac{ \sin(x) - \sin(a) }{ x - a } x > a\]

Answer 19

\[5n + 1 \le 7\]
\[5n \le 6\]
\[n \le \frac{ 6 }{ 5 } (1.2) \]
n = 1
r! = 5 * 1 + 1 r! = 3 ! r = 3
but I am not sure if assuming 5n + 1 <= 7 is right or wrong. teoretically it is as r <= 7