Between 1 and 30, each number with same last digit occurs 3 times.
i) 30C1, Last digit is 0..
ii) 27C1, Last digit is 1..
iii) 24C1, Last digit 2...
iv) 21C1,
v) 18C1,
vi) 15C1,
vii) 12C1,
Possible outcomes: 30C7
Favorable outcomes: 30C1.27C1.24C1.21C1.18C1.15C1.12C1
Probability: 30C1.27C1.24C1.21C1.18C1.15C1.12C1/30C7
Is my answer correct?
Your calculation is on the right track, but there are some errors and oversights. It yields the number of ways to pick one card, then a card with a different last digit, then a third card with yet another last digit, and so on. Thus, you’re counting each set of $7$ cards $7!$ times, once for each of the $7!$ orders in which it could have been dealt. In your denominator, however, you’re just counting sets of $7$ cards, without regard for order, and you’re counting sets of $7$ cards from a deck of $52$ cards, not $30$. You must correct the $52$ to $30$ and either divide the numerator by $7!$ to get rid of the overcounting or, equivalently, multiply the denominator by $7!$ to take into account the order in which each hand of $7$ cards is dealt. When you do that, you get
$$\begin{align*} \frac{\binom{30}1\binom{27}1\binom{24}1\binom{21}1\binom{18}1\binom{15}1\binom{12}1}{7!\binom{30}7}&=\frac{30\cdot27\cdot24\cdot21\cdot18\cdot15\cdot12}{7!\binom{30}7}\\ &=\frac{3^7\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4}{7!\binom{30}7}\\ &=\frac{3^7\cdot10!}{3!7!\binom{30}7}\\ &=\frac{3^7\binom{10}7}{\binom{30}7}\;. \end{align*}$$
There is a shorter way to arrive at this result. There are $10$ possible last digits, so there are $\binom{10}7$ ways to choose $7$ of them to be the last digits of a ‘good’ set. There are $3$ cards having each of those $7$ last digits, so once we know what the $7$ last digits are, there are $3^7$ ways to choose the specific $7$ cards having those last digits. Thus, there are $3^7\binom{10}3$ possible ‘good’ hands.