Discrete maths

Question

Number of ways to distribute 6 distinct balls into 3 distinct boxes,such that each box contains at least 1 ball?

Comments

Actual answer is 180. Can refer over here

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

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I am getting 3240.

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Hi Shubhanshu, I think we should apply S(k,n) × n!  formula for this question. Don't you think? Formula is mentioned in the link shared by me.

Answer 1

This problem can be solved by counting the number of ONTO functions from A->B
where |A|=6, and |B|=3

# ONTO Functions = Total functions - {Not Onto Functions}
=$3^6 - (3C1*2^6 - 3C2*1^6 + 3C3*0^6)$
= 729 - (192 - 3 + 0)
= 729 - 189 = 540

Comments

$\left | A \right | = 6$

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Typo!, corrected.

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Hi manu, can you please explain briefly how you calculated not onto functions?

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|A| = m and |B| = n and 1<=n<=m

Number of onto functions = ∑ (-1)^{n-r  n}C_r r^m   r vary from 1 to n

= 3C1 (1)⁶  - 3C2 (2)⁶ + 3C3 (3)⁶

= 540

Answer 2

6 distinct balls , 3 distinct boxes     // condition : each box must be non-empty

part 1) total ways to distribute 6 distinct balls to 3 distinct boxes , when there is no condition

every ball has 3 homes and there are 6 such balls so it's 3*3*3*3*3*3=3⁶

Out of these 3⁶ ways we have to subtract ways when there is exactly one box is empty ans when there is exactly 2 boxes are empty.

Part 2 ) case 1 : # ways when exactly 2 boxes are empty :  select 2 boxes out of 3 (3C2)and then put 6 balls in one box(1⁶)

total = 3C2*1⁶=3 ways

Part 2 ) case 2 : # ways when exactly 1 box is empty :-select one box out of 3 (3C1 ways) put 6 balls into 2 boxes such that no box remain empty (let consider this x)

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now we have to find 'x' , follow same step total ways - when exactly one empty = 2⁶-(2C1)*1⁶=62 ways

put the value of x so # ways when exactly 1 box is empty =(3C1)*62=186

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so total ways when no box would be left empty = (part 1)- (part 2 case 1)-(part 2 case 2)

3⁶-3-186=729-189=540 ways

Answer 3

one more way :)