Find the number of ways in which 4 letters can be selected from the 9 letters of the word ALGORITHM, given at least one vowel is included. My solution:
3 cases:
1 vowel:
$6P3 \times 3C1 = 360 $
2 vowels:
$6P2 \times 3C2 \times 2! = 180 $
3 vowels:
$6P1 \times 3! = 36 $
Total number of ways: 360 + 180 + 36 = 576
Am I correct?
 A: Taking the header to be right, your answer is wrong !
Do pay heed to the exact wordings of questions.
It reads "... $4$ letters can be selected ...."
There is a difference between selection and arrangement
Btw, even had the question been about arrangements, your answer would have been wrong.
Added Guidance
Assuming it was about arrangements, the first case would have been $^6C_3\times^3C_1\times 4!$
And if it was about selections, just $^6C_3\times^3C_1$
As a final word of guidance for at least types of questions, it is better to work with the complement of the desired event, so if we want, for example, P(at least one vowel), compute  $1-$ P(no vowel)
A: Using the Inclusion-Exclusion principle, we count the number of combinations that use $1$ vowel, subtract those that use $2$, and add those that use all $3$.
The generating function for this is
$$[x^4]:(1-(1-x)^3)\frac{1}{(1-x)^6}$$
where
$$1-(1-x)^3=\binom{3}{1}x-\binom{3}{1}x^2+x^3$$
$$\frac{1}{(1-x)^6}=\binom{5}{5}+\binom{6}{5}x+\binom{7}{5}x^2+\binom{8}{5}x^3+\dots$$
and gives $3\cdot56-3\cdot21+6=111$.
Also, the GF can be re-arranged to give
$$[x^4]:\frac{1}{(1-x)^6}-\frac{(1-x)^3}{(1-x)^6}$$
$$[x^4]:\frac{1}{(1-x)^6}-\frac{1}{(1-x)^3}$$
$$=\binom{9}{5}-\binom{6}{2}$$
$$=111$$
