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

1.  Repetition is not allowed and the order of picking matters?
2.  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.
edited | 369 views
0
what if

(i) Repetition is  allowed and the order of picking matters?

(ii) Repetition is not allowed and the order of picking  matter?
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what is the meaning of repetition here ...any one please?
+2

Let set S={1,2,3,4,5}

// we can pick any element any number of times.

a) repetition is not allowed  and order of picking matters.

repetition not allowed  means we can't pick any element more than one time in a single extraction of elements.

like we have to pick 2 elements then those can be (1,2)(2,1)(1,3)(3,1)(1,4)(4,1)(1,5)(5,1)(2,3)(3,2)(2,4)(4,2)(2,5)(5,2)(3,4)(4,3)(3,5)(5,3)(4,5)(5,4) so 20 ways  (nCr)*r!

b)repetition is allowed and order of picking doesn't matter.

Here repetition is allowed so (1,1)(2,2)(3,3)(4,4)(5,5) can we out of those 2 we're picking.here order doesn't matter means (1,2) and (2,1) is same. so total=10+5=15 ways.

c)repetition is allowed and order of picking matters.

so in this case simply there would be n choices for each place out of r => nr

d) repetition is not allowed but order of pickings doesn't matter.

(1,2)(1,3)(1,4)(1,5)(2,3)(2,4)(2,5)(3,4)(3,5)(4,5)

simply nCr ways

+1
good explanation ...

(i) Repetition is not allowed and the order of picking matters $=$r arrangement with no repetition

$=^{n}\!\text{P}_{r} = \frac{n!}{(n-r)!}$

(ii) Repetition is allowed and the order of picking does not matter $=$ combination with unlimited repetition

$=^{\left(n-1+r\right)}\!\text{C}_r=\frac{n-1+r!}{(n-1)!r!}$
Hence, ans is Option A.
edited
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It will work in case anyone dont understand....

+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.