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In how many ways we can put $n$ distinct balls in $k$ dintinct bins??

Will it be $n^{k}$ or $k^{n}$?? Taking example will be easy way to remove this doubt or some other ways possible??

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Suppose if we have 5 balls 1,2,3,4,5

and 6 bins a,b,c,d,e,f

So first we select a random ball -> 5 ways (either 1 or 2 or 3 or 4 or 5) lets say we selected ball 2.

then we select a random bin to put the ball 2 in it -> 6 ways(either a or b or c or d or e or  f) lets say we selected bin c

so total ways = 5*6 = 30

next we select a random ball -> 4 ways (either 1 or 3 or 4 or 5) lets say we selected ball 3.

then we select a random bin to put the ball 3 in it -> 6 ways(either a or b or c or d or e or  f)  repetition is allowed

so total ways = 4*6 = 24

.......

Repeat the steps for remaining balls

so total ways = (5*6)*(4*6)*(3*6)*(2*6)*(1*6) =$5! * 6^{5}$ = $n! * k^{n}$

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that is not correct
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I calculated permutation but in this order will not matter so we will remove n! and so answer will be $k^n$.

For ball $1$ we can select any of the $6$ boxes. = $6$ ways

For ball $2$ we can select any of the $6$ boxes. = $6$ ways

For ball $3$ we can select any of the $6$ boxes. = $6$ ways

For ball $4$ we can select any of the $6$ boxes. = $6$ ways

For ball $5$ we can select any of the $6$ boxes. = $6$ ways

So total ways = $6*6*6*6*6$ = $6^{5}$ ways = $k^n$.
+1
For $1^{st}$ ball there are $k$ options .

For $2^{nd}$ ball there are $k$ options .

..

.

For $n^{th}$ ball there are $k$ options .

Therefore total ways is $k^n$.
+2
Each ball has k choices in which it can be put into a bin.Hence $k^n$
+1

yes , Perfect ans @Ayush Upadhyaya

and why do we choice for ball, not for bag??

+1

@srestha-Take an example.say we have 5 balls and 3 bags.A good wrong answer is $5^3$.Each bag has 5 choices to choose a ball.But then the flaw is when Bag1 has choosen ball 1 say, we are remaining with 4 balls and thus remaining bags won't have 5 choices each.

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yes, right.

Suppose if we have $5$ balls $1,2,3,4,5$ and $4$ bins $a,b,c,d$ i.e. $k=4$ and $n=5$

For ball $1$ we can select any of the $4$ boxes. = $4$ ways

For ball $2$ we can select any of the $4$ boxes. = $4$ ways

For ball $3$ we can select any of the $4$ boxes. = $4$ ways

For ball $4$ we can select any of the $4$ boxes. = $4$ ways

For ball $5$ we can select any of the $4$ boxes. = $4$ ways

So total ways = $4*4*4*4*4$ = $4^{5}$ ways = $k^n$ ways

$\therefore$ We can put $n$ distinct balls in $k$ distinct boxes in $k^n$ ways.
by Boss (21.6k points)
selected
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CASE 1 :  n distinct balls , k distinct bins .

here,  every bin can contain zero ball, one ball, two balls,......n balls etc. every ball has k choices to reside into the bin.

so , total ways here = k*k*k*...n times = k^n

Case 2 : In number system , n symbols (i.e. radix n)  k digit

In every position , n symbols can repeat. so in every digit position has  n choices. So, n*n*n*....k times = n^k

Difference between two cases ---> in case 1 , k  can repeat , case 2 ---> n can repeat. so total ways = (the number repeat) ^ how many times repeat .

by (293 points)
it will (k-1+n)Cn    i.e     (k-1+n)!/(k-1)!n!
by (11 points)

+1 vote