Say you have the usual 52 card deck, 13 ranks and 4 suits. If I have a hand of 6 cards, how many hands are there where at least half the cards have an odd number as rank? they can be of any suit.

This is what I'm doing: So 13 ranks (assume they're numbered from 1 to 13 and the ace is a one) , this means that there are 7 odd ranks (1, 3, 5, 7, 9, 11, 13). We want half meaning at least 3 of the 6 cards in my hand need to be odd. So:

I want the cases where there 3 odd cards + 4 odd cards + 5 odd cards + 6 odd cards. Is this how to represent it? $7\choose3 $ + $7\choose4 $ + $7\choose5 $ + $7\choose6 $ I feel like I'm missing something

Say you have the usual $52$ card deck, $13$ ranks and $4$ suits. If I have a hand of $6$ cards, how many hands are there where at least half the cards have an odd number as rank? they can be of any suit.

This is what I'm doing:

So $13$ ranks (assume they're numbered from $1$ to $13$ and the ace is a one) , this means that there are $7$ odd ranks ($1$, $3$, $5$, $7$, $9$, $11$, $13$). We want half meaning at least $3$ of the $6$ cards in my hand need to be odd. So:

I want the cases where there $3$ odd cards $+ 4$ odd cards $+ 5$ odd cards $+ 6$ odd cards. Is this how to represent it?

$$\frac 73 + \frac74 + \frac 75 +\frac76 $$

I feel like I'm missing something