Can we solve this only by using bijection?

Question

The question is :

The number of possible outcomes in a throw of n ordinary dice in which at least once of the dice shows an odd number are:

Now, we can simply apply bijection principle , and calculate the required number of ways as follows:

Required Number of ways = Total number of outcomes - No. of outcomes in which there is no odd number \
= $6^n$-$3^n$

However I tried to do this question by another method

Since we require at least one dice to give us odd number (i.e.3 possible ways) , we can consider cases which are exactly one dice showing odd digits, exactly two die showing odd digits,exactly 3 die showing odd digits.... and so on till *exactly all the die give us odd digits

After we calculate the total possible ways in each case and add them , we can use binomial expansion formulaes to condense the answer in exponential forms ( with base = some constant and exponent = n)

I tried this method , and this is how it goes :-

First Case = \
One dice gives me odd digits \\ = $3*6*6*6*6*6 .... n-1\ times$ = $3.6^{n-1}$ \
Similarly the other cases are calculated in the same way , and when I add them I receive this : \
$3.6^{n-1}$ + $3^2.6^{n-2}$ + $3^3.6^{n-3}$ + ..... $3^n.6^0$\
Clearly this can be expressed in the form of binomial expression as = $(3+6)^n$ - $3^n$\
= $9^n - 6^n$, which is not the correct answer

What am I doing wrong here?

Answer 1

For $~k \in \{0,1,2,\cdots,n\},~$ the number of ways of throwing exactly $~n~$ dice, with exactly $~k~$ of them even is

$$\binom{n}{k} 3^k \times 3^{n-k} = \binom{n}{k} \times 3^n. \tag1 $$

In the LHS of (1) above, the first factor refers to the number of distinct (equally likely ways) of selecting which $~k~$ dice will show an even number.

Then, with the specific dice that will show even specified, the number of ways that you can have exactly $~k~$ dice showing an even number, and $~(n-k)~$ dice showing an odd number is $~3^k \times 3^{n-k}.$

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So, your alternative approach should be

$$\sum_{k=1}^n \left[ ~\binom{n}{k} \times 3^n ~\right] = 3^n \times \sum_{k=1}^n \binom{n}{k}. \tag2 $$

The RHS of (2) above can be simplified by noting that

$$\sum_{k=0}^n \binom{n}{k} = 2^n.$$

Therefore, the expression in (2) above equals

$$3^n \times \left[ ~2^n - \binom{n}{0} ~\right] = 3^n \times \left[ ~2^n - 1 ~\right] = 6^n - 3^n.$$