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In how many different ways can $r$ elements be picked from a set of $n$ elements if

(i) Repetition is not allowed and the order of picking matters?

(ii) Repetition is allowed and the order of picking does not matter?

  1. $\frac{n!}{\left(n - r\right)!}$ and $\frac{\left(n + r - 1\right)!}{r!\left(n - 1\right)!}$, respectively.
  2. $\frac{n!}{\left(n - r\right)!}$ and $\frac{n!}{r!\left(n - 1\right)!}$, respectively.
  3. $\frac{n!}{r!\left(n - r\right)!}$ and $\frac{\left(n - r + 1\right)!}{r!\left(n - 1\right)!}$, respectively.
  4. $\frac{n!}{r!\left(n - r\right)!}$ and $\frac{n!}{\left(n - r\right)!}$, respectively.
  5. $\frac{n!}{r!}$ and $\frac{r!}{n!}$, respectively.
asked in Combinatory by Veteran (39.7k points) 253 1302 1929
retagged by | 222 views

2 Answers

+10 votes
Best answer

(i) Repetition is not allowed and the order of picking matters =
      = r arrangement with no repetition = npr = $\frac{n!}{(n-r)!}$

(ii) Repetition is allowed and the order of picking does not matter = 
      = combination with unlimited repetition = n-1+rCr = $\frac{n-1+r!}{(n-1)!r!}$
Option A

answered by Veteran (15.3k points) 17 51 129
selected by
+1 vote

Order of picking matters means,(1). a,b,c (2)a,c,b  ,both (1) and (2) are treated as different.

Order of picking does not matters means,(1). a,b,c (2)a,c,b  ,both (1) and (2) are treated as same.

(i) Repetition is not allowed and the order of picking matters = C(n , r) ⨉ P(r , r)
      =   C(
n , r) ⨉ r! = P(n , r) = n! / (n-r)! = It is eqt to say Permutation of r objects in the pool of n object.

(ii) Repetition is allowed and the order of picking does not matter =  C(n+r-1 , r)

    = (n+r-1)! / r!(n-1)!  = It is eqt to say the combination of r objects in n objects with unlimited repetition is allowed.

The correct answer is (A).

answered by Boss (8.8k points) 3 8 12
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