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?

  • $\begingroup$ Are you sure it's of the form of binomial expressions? I don't see any $ { n \choose k } $ terms. It looks more like a GP with common ratio 1/2? $\quad$ Also $(3+6)^n - 3^n \neq 9^n - 6^n$. $\endgroup$
    – Calvin Lin
    Feb 1 at 22:32
  • 1
    $\begingroup$ Your attempt is hard to follow because you seem to have left out the very important binomial coefficients and also written $-3^n$ instead of $-6^n$. Assuming those are just typos, your logical error is that you are double counting. For example, you counted the result 1122222 three times (twice as "one odd die" and once as "two odd dice"). To fix this, you can use the inclusion-exclusion principle. $\endgroup$
    – Karl
    Feb 1 at 22:33
  • 1
    $\begingroup$ A reliable way to find your own mistakes on counting problems is to test your steps on concrete examples with smaller numbers. $\endgroup$
    – Karl
    Feb 1 at 22:39

1 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 $~3^k \times 3^{n-k}.$

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

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


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