Self Doubt-Combinatory

Question

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

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

Comments

yes , Perfect ans @Ayush Upadhyaya

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

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@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.

Answer 1

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.

Comments

For other formulaes i found this useful

http://www.careerbless.com/aptitude/qa/permutations_combinations_imp7.php

Answer 2

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 . 

Answer 3

it will (k-1+n)Cn    i.e     (k-1+n)!/(k-1)!n!