answer 90 or 444?

Question

Answer 1

6 different pencils in 3 different bags can be placed in 540 ways

$3^6\ -\ ^3C_12^6\ +\ ^3C_21^6 = 729-192+3 =540$

All bags can't have same balls means 2,2,2 should not occur

$\frac{6!}{2! \times 2! \times 2!} = 90$

So, ans is 540 - 90 = 450 

Comments

You can get 540 using #onto functions from 6 diff elements to 3 diff elements.

---

@yash see this will clear your doubt

theorem 5 --2)

http://cns-web.bu.edu/~eric/EC500/attachments/ON(2d)LINE(20)READINGS/ballsinboxe.pdf

---

@Lokesh. I even tried it this way.

case1: 1 pencil in 1st box

This 1 pencil can be choosen in 6C1 ways.

For remaining 5 pencils,  #onto functions from 5 different balls to 2 different boxes. 

So, total = 6C1 *  S(5,2) * 2! = 180

case 2: 3 pencils in 1st box

These 3 pencils can be choosen in 6C3 ways.

Now, for remaining 3 pencils, #onto functions from 3 diff pencils to 2 diff bags = s(3,2) * 2!

Thus, total = 6C3 * S(3,2) * 2! = 120

case 3: 4 pencils in 1st box

These 4 pencils can be choosen in 6C4 ways

For remaining 2 pencils, #onto functions from 2 diff pencils to 2 diff bags = S(2,2) * 2!

Thus, total = 6C4 * S(2,2) * 2! = 30

case 4: 2 pencils in 1 st box

This can be done in 6C2 ways.

We have 3 subcases:

a) 2 pencils in 2 nd box and 2 pencils in 3rd box

b) 1 pencil in 2n bag and 3 pencils in 3rd bag

    This accounts to 4C1*3C3  = 4

c) 3 pencils in 2nd bag and 1 pencil in 3rd box

    This account to 4C3 * 1C1 = 4

Thus, total = 6C2 * (4+4) = 120

Sum of all cases = 180 + 120 + 30 + 120 = 450

This was the same approch as you did  but slightly in different way. So, 450 should be correct.

Answer 2

Six pencils can be put in three bags such that no bag contains the same number of pencils, only if:-

Case 1: One bag contains 1 pencil, the other one contains 2 pencils and the other one contains 3 pencils.

Case 2: One bag contains 1 pencil, the other one contains 1 pencil and the other one contains 4 pencils.

The pencils are different.

Analysis of Case 1:

3 pencils can be chosen from 6 in $\binom{6}{3}$ ways = 20 ways.

2 pencils from remaining 3 can be chosen in $\binom{3}{2}$ ways = 3 ways.

1 pencil from the remaining 1 pencil can be chosen in 1 way.

Now that we have formed the groups, the pencils can be arranged in 3 bags in 3! = 6 ways

So, total ways = 20 * 3 * 1 * 6 =360 ways.

Analysis of Case 2:

4 pencils can be chosen from 6 in $\binom{6}{4}$ ways = 15 ways.

1 pencils from remaining 2 can be chosen in $\binom{2}{1}$ ways = 2 ways.

1 pencil from the remaining 1 pencil can be chosen in 1 way.

Now that we have formed the groups, the pencils can be arranged in 3 bags in 3! = 6 ways

So, total ways = 15 * 2 * 1 * 6 =180 ways.

By addition rule of counting, number of ways = 360 + 180 = 540

Comments

all bags cannot have same balls...subtract this

$\frac{6!}{2! \times 2! \times2!} = 90$

ans = 540 - 90 = 450

---

I have not taken that case of all bags having same pencils in the solution

---

for case 2:
step 1: choose 4 balls $\binom{6}{4}$
step 2: put in one of the bag $\binom{3}{1}$
step 3: remaining two balls can be put in 2 ways only (1,2) or (2,1)

so 15*3*2 = 90

same goes for case 1 though your ans matches