Solve:

Is this permutation and combination?

ABCthings:
Is this permutation and combination?

Yes

No body to solve it?

Lol. This is very simple na. Check out the maths challenges I gave and see if you can solve it. If no mathematician on nairaland can solve it, I will post the solution in 3 days time. Try your luck.

Before solving here are things to note:
1) P(n, r) = (n!)/(n - r)! , 2) C(n, r) = (n!)/ [ r!(n - r)! ]
3) 0! = 1! = 1 and
4) n! = n(n - 1)(n - 2)(n - 3)... × 3 × 2 × 1, n is an integer & n ≥ 2.

With these, let's get on with it.

1) To show (n + 1) P(n, n) = P(n+1, n)

FROM LHS :
(n + 1) P(n, n) = (n + 1) [(n!)/(n - n)!]
= [(n + 1)(n!)] / (n - n)! = (n + 1)! / (n - n)!
since (n - n)! = 0! = 1 = 1! , you have (n - n)! = 1! = [(n + 1) - n]!
[(n + 1)!/(n - n)! ] = [ (n + 1)! / [(n+1) - n]! ]
= P(n+1, n)

FROM RHS :
P(n+1, n) = (n + 1)! / [(n+1) - n]!
since [(n+1) - n]! = 1! = 0! = (n - n)! ,
= ( [(n + 1)(n!)] / [(n+1) - n]! )
= (n + 1) [(n!)/(n - n)!] = (n + 1) P(n, n)

∴ (n + 1) P(n, n) = P(n+1, n)

2) to show that P(n, r+1) = (n - r) P(n, r)

FROM LHS :
P(n, r+1) = (n!) / [n - (r+1)]!
multiplying and dividing by (n - r)!
= [(n - r)(n!)] / [ (n - r)[n - (r+1)]! ]
= [(n - r)(n!)] / (n - r)! = (n - r)[(n!)/(n - r)!]
= (n - r) P(n, r)

FROM RHS : simply use the above workings to work backwards to show (n - r) P(n, r) can be rewritten as P(n, r+1)

3) To show that P(n, r) = (r!) C(n, r)

FROM LHS :
P(n, r) = (n!)/(n - r)!
multiplying and dividing by r! ,
P(n, r) = [(r!)(n!)] / [(r!)(n - r)!]
= (r!) [ (n!)/ [(r!)(n - r)!] ] = r! C(n, r)

FROM RHS :
C(n, r) = (n!) / [(r!)(n - r)!]
r! C(n, r) = [(r!)(n!)] / [(r!)(n - r)!] = (n!)/(n - r)!
r! C(n, r) = P(n, r)

∴ P(n, r) = r! C(n, r)

4) To show that C(n, r) = C(n, n - r)

FROM LHS :
C(n, r) = (n!) / [(r!)(n - r)!]
r! = [ n - (n - r) ]!
(n!)/ [(r!)(n - r)!] = (n!) / ( [n - (n - r)]!(n - r)! )
if d = (n - r)
(n!) / ( [n - (n - r)]! (n - r)! ) = (n!) / [(n - d)!(d!)]
= C(n, d) = C(n, n - r)

FROM RHS :
C(n, n - r) = (n!) / [ (n - r)! [n - (n - r)]! ]
n - (n - r) = r
(n!) / [ (n - r)! [n - (n - r)]! ] = (n!) / [(r!)(n - r)!]
= C(n, r)

∴ C(n, r) = C(n, n - r)

5) To show C(n, r) + C(n, n - r) = C(n+1, r)

I've shown in the previous proof that C(n, r) = C(n, n - r) . Hence

C(n, r) + C(n, n - r) = 2 C(n, r) = 2 C(n, n - r) = C(n+1, r)
∴ on simplification,
C(n+1, r) = 2 C(n, r) = [(2)(n!)] / [(r!)(n - r)!]

We begin with C(n+1, r) . Observe that

C(n+1, r) = (n + 1)! / [ (r!)(n + 1 - r)! ]
= [ (n + 1)(n!) ] / [ (r!)(n + 1 - r)(n - r)! ]
= [ (n + 1)/(n + 1 - r) ] × [ (n!) / [(r!)(n - r)!] ]
= [ (n + 1)/(n + 1 - r) ] C(n, r)

Since [ (n + 1)/(n + 1 - r) ] ≠ 2 for any arbitrary choices of n and r , we can say that the statement C(n, r) + C(n, n - r) = C(n+1, r) is invalid except for n and r choices that simultaneously satisfy the linear equation n - 2r + 1 = 0

Sl2348, I have completed your assignment.

Martinez39:




Sl2348, I have completed your assignment.

Thanks bro, I really appreciate.