Selection of at least one vowel and one consonant 
Find the number of ways of selection of atleast one vowel and one consonant from the word TRIPLE.

Attempt: 
We have $(I,E)$ and $(T,R,P,L)$
We choose $\dbinom{2}{1} \times \dbinom{4}{1}\times 2^4$
Explanation of attempt:
First choose one vowel.
Then choose one consonant.
Then the rest have 2 choices each - to be selected or not.
Why is my method incorrect? 
(The answer is $45$)
 A: You can choose one vowel or two vowels ( so j goes from 1 to 2) and take also one consonant or more so i (number of consonants) goes from 1 to 4. So we have :
$$1\le j\le 2$$
$$1\le i\le 4$$
$$P=\sum_{j=1}^2\binom 2 j \sum_{i=1}^4\binom 4 i=(\binom 2 1 +\binom 2 2 ) \sum_{i=1}^4\binom 4 i=3\sum_{i=1}^4\binom 4 i=3*15=45$$
A: I think there's a simpler way of seeing this than the existing answers: you need to pick a non-empty subset of $\{I, E\}$ and a non-empty subset of $\{T,R,P,L\}$. A set of $n$ elements has $2^n - 1$ non-empty subsets, so the solution is $(2^2 - 1)(2^4 - 1) = 3 \times 15 = 45$.
A: Since there are 6 letters, the total number of possible ways is $2^6=64$
We can simplify the question by finding the ways that fail:  All consonants or all vowels.  All consonants is $2^4=16$.  All vowels is $2^2=4$.
Therefore, total number of ways is $64-(16+4–1)=45$.
In your attempt, you count the possibilities.  However, notice that you have repeated answers.  If the question asks for 2 letters, 1 vowel and 1 consonant, it is simply $^2C_1*^4C_1$.  However, you can choose more letters here.  You can choose another set of vowel and consonant.  Now that repeats.
A: Lets break down the problem.
We need to find the number of ways of selecting at least 1 vowel and 1 consonant. 
We have two vowels.
So we can select one vowel or  two vowels.
This can be done in 2C1 + 2C2 = 2 + 1 = 3 Ways.
We have 4 consonants.
We can select 1,2,3 or 4 Consonants.
4C1 + 4C2 + 4C3 + 4C4 = 4 + 6 + 4 + 1 = 15 Ways.
Number of ways of selecting at least one vowel and one consonant is 3 * 15 = 45.
