Question

Prove that if r ≤ s ≤ n , P ( n , s) is divisible by P ( n , r)

Answer 1

$\large\underline{\sf{Solution-}}$

We know,

$\rm \: P(n,r) = \dfrac{n!}{(n - r)!}$

can be further rewritten as

$\rm \: P(n,r) = \dfrac{n(n - 1)(n - 2) - - - - (n - r + 1)(n - r)!}{(n - r)!}$

$\rm \: P(n,r) = n(n - 1)(n - 2) - - - - (n - r + 1)$

Now, as it is given that

$\rm \: r \leqslant s \leqslant n$

Let assume that

$\rm \: s = r + k \: where \: k \: \in \: [0, \: s - r]$

Now, Consider

$\rm \: P(n,s)$

$\rm \: = \: \dfrac{n!}{(n - s)!}$

$\rm \: = \dfrac{n(n - 1)(n - 2) - - - - (n - s + 1)(n - s)!}{(n - s)!}$

$\rm \: = n(n - 1)(n - 2) - - - - (n - s + 1)$

On substituting the value of s, we get

$\rm \: = n(n - 1)(n - 2) - - - - (n - r - k + 1)$

$\rm \: = n(n - 1)(n - 2) - - - - [n - (r + k - 1)]$

can be further rewritten as

$\rm \: = n(n - 1)(n - 2)... [n - (r - 1)](n - r)[n - (r + 1)]... [n - (r + k - 1)]$

$\rm \: = \red{n(n - 1)(n - 2)... [n - (r - 1)] }(n - r)[n - (r + 1)]... [n - (r + k - 1)]$

$\rm \: = \red{P(n,r) }(n - r)[n - (r + 1)]... [n - (r + k - 1)]$

$\rm\implies \: P(n,s)= \red{P(n,r) }(n - r)[n - (r + 1)]... [n - (r + k - 1)]$

$\rm\implies \: P(n,s) \: is \: divisible \: by \: P(n,r)$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{\rm{  \:C(n,r) = \frac{n!}{r! \: (n - r)!} \: }}$

$\boxed{\rm{  \:P(n,r) \: = \: r! \: C(n,r) \: }}$

$\boxed{\rm{  \:C(n,x) = C(n,y) \: \rm\implies \:x = y \: \: or \: \: n = x + y \: }}$

$\boxed{\rm{  \: \frac{C(n,r)}{C(n,r - 1)} \: = \: \frac{n - r + 1}{r} \: }}$

$\boxed{\rm{  \:C(n,r) + C(n,r - 1) = C(n + 1,r) \: }}$

$\boxed{\rm{  \:P(n,n) \: = \: n! \: }}$

$\boxed{\rm{  \:C(n,n) \: = \: C(n,0) \: = \: 1 \: }}$

$\boxed{\rm{  \:C(n,n - 1) \: = \: C(n,1) \: = \: n \: }}$