another P & C Q (1 Viewer)

[5647382910]

sorry to ask for help yet again....... this stuff just doesnt click with me

from the DEFINITION of nPr and nCr proove:
i) n+1^P_r = nPr + r. (n^P_r-1)
ii) nCr = ((n - r + 1)/r).(n^C_r-1)

thanks in advance

 

[dp624]

from (N+1) objects you are to choose and arrange (r) of them. Assume r<N
now, let's separate them into a group of (N) and (1) Objects
There are two ways you can choose and arrange (r) objects from that
(r-1) from (N) and (1) from (1) OR (r) from (N) and 0 from the (1)

The first case is ^NP_r-1 - you choose and arrange r-1 items. Then you have to place the last one, which can go into (r) places. Thus r(^NP_r-1 )

The second case is equal to ¹P₀ multiplied by ⁿP_r, which is ⁿP_r.

So for part i) there are r(^NP_r-1) + ⁿP_r ways.

for part ii)
you are CHOOSING r items from a selection of n.
now, let's say you chose (r-1) items from n. There would be ⁿC_r-1 ways to do so.
Now to choose the last item. There are n- (r-1) items left -> ie. (n-r+1)
So you multiply the initial by this. then divide by (r) to unorder the selection. So it's (n-r+1)/r * ⁿC_r-1

unless i understood his definition wrong... it'd be MUCH easier to do it by N! over (n-r)! r! haha

 

[Trebla]

sorry to ask for help yet again....... this stuff just doesnt click with me

from the DEFINITION of nPr and nCr proove:
i) n+1^P_r = nPr + r. (n^P_r-1)
ii) nCr = ((n - r + 1)/r).(n^C_r-1)

thanks in advance

ⁿP_r = n! / (n - r)!
ⁿC_r = n! / r!(n - r)!

i) RHS = ⁿP_r + r.ⁿP_{r - 1}
= n! / (n - r)! + r.n! / (n - (r - 1))!
= n! / (n - r)! + r.n! / (n - r + 1)!
= n! / (n - r)! [1 + r / (n - r + 1)]
= n! / (n - r)! [(n - r + 1 + r) / (n - r + 1)]
= n! / (n - r)! [(n + 1) / (n - r + 1)]
= (n + 1)! / (n + 1 - r)!
= ^{n + 1}P_r
= LHS

ii) RHS = ⁿC_{r - 1} (n - r + 1) / r
= {n! / [(r - 1)!(n - (r - 1))!]}{(n - r + 1) / r}
= {n! / [(r - 1)!(n - r + 1)!]}{(n - r + 1) / r}
= {n! / [(r - 1)!(n - r)!(n - r + 1)]}{(n - r + 1) / r}
= n! / r.(r - 1)!(n - r)!
= n! / r!(n - r)!
= ⁿC_r
= LHS

 

[dp624]

ⁿP_r = n! / (n - r)!
ⁿC_r = n! / r!(n - r)!

i) RHS = ⁿP_r + r.ⁿP_{r - 1}
= n! / (n - r)! + r.n! / (n - (r - 1))!
= n! / (n - r)! + r.n! / (n - r + 1)!
= n! / (n - r)! [1 + r / (n - r + 1)]
= n! / (n - r)! [(n - r + 1 + r) / (n - r + 1)]
= n! / (n - r)! [(n + 1) / (n - r + 1)]
= (n + 1)! / (n + 1 - r)!
= ^{n + 1}P_r
= LHS

ii) RHS = ⁿC_{r - 1} (n - r + 1) / r
= {n! / [(r - 1)!(n - (r - 1))!]}{(n - r + 1) / r}
= {n! / [(r - 1)!(n - r + 1)!]}{(n - r + 1) / r}
= {n! / [(r - 1)!(n - r)!(n - r + 1)]}{(n - r + 1) / r}
= n! / r.(r - 1)!(n - r)!
= n! / r!(n - r)!
= ⁿC_r
= LHS

so, Trebla, is the definition of nCr the ways to choose r from a set of n? Or is it the algebra equation?
I was taught the former

 

[jet]

The definition is the algebraic equation. This is just cause it came from the binomial theorem first, and the link to combinations came later.

 

[5647382910]

thanks for the help guys;
would they ask for a proof from a definition in an exam?

 

[jet]

Sometimes. definitely in 4u.

Just remember the definition and work from there.