atleast two balls identical in 5 bags There are five bags each containing identical sets of ten distinct chocolates. One chocolate is picked from each bag.
The probability that at least two chocolates are identical is?
I know the solution can be arrived by using 1-(none ball is picked is same) like this P(No two chocolates are identical) = $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 30240$
so P(no two chocolates are identical) $ = \dfrac{30240}{100000} = 0.3024$
P(At least two chocolates are identical) = $ 1$ – P(No two chocolates are identical)
$ = 1 – 0.3024 = 0.6976$
But I used this approach
probability at least $2$ bags have identical chocolates is equal to = probability of two exactly two bag having same chocolates + probability of exactly three bag having same chocolates +probability of exactly four bags having same chocolates + probability of five bags have identical chocolates.
Now I have calculated two bag having same chocolates is calculate as
No. of ways of choosing two bags $C(5,2)$ and the these two bags have same chocolates rest have different chocolates so each $C(5,2)$ will have $10 \cdot 9 \cdot 8 \cdot 7$ (basically reducing number of bags having distinct chocolates to 4 by grouping pair )
So number of bags having exactly two identical chocolates is $ \displaystyle {5 \choose 2} \cdot 10 \cdot 9 \cdot 8 \cdot 7 = 50400$
Similarly number of bags having exactly three identical chocolates $ \displaystyle {5 \choose 3} \cdot 10 \cdot 9 \cdot 8 = 7200$
Similarly number of bags having exactly four identical chocolates $ \displaystyle {5 \choose 3} \cdot 10 \cdot 9 = 450$
Similarly number of bags having exactly five identical chocolates = $ \displaystyle {5 \choose 5} \cdot 10 = 10$
Total $58060$
out of total configuration $10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 =100000$
So answer becomes $ \dfrac {58060}{100000} =0.5806$
I know I am missing something but what can you please help
 A: $58060$ counts all outcomes where there is an identical chocolate drawn from two or more bags and we draw different chocolates from all remaining bags. But you have not counted outcomes where there are two sets of identical chocolates drawn:
case $1$: Out of $5$ bags, we draw an identical chocolate from two bags and another identical chocolate from another two bags.
case $2$: Out of $5$ bags, we draw an identical chocolate from two bags and another identical chocolate from remaining three bags.
Case $1$: Choose $3$ chocolate types, choose which $2$ repeat and then arrange $5$ chocolates to be drawn from $5$ bags.
$ \displaystyle {10 \choose 3} {3 \choose 2} \cdot \frac{5!}{2! ~ 2!} = 10800$
Case $2$: Choose $2$ chocolate types, choose which type we have three chocolates of and then arrange $5$ chocolates to be drawn from $5$ bags.
$ \displaystyle {10 \choose 2} {2 \choose 1} \cdot \frac{5!}{3! ~ 2!} = 900$
So the desired probability is
$ \displaystyle = \frac{58060 + 10800 + 900}{10^5} = 0.6976$
