The World Almanact features a "perpetual calendar," a collection of $14$ possible calendars. Does this suffice to ensure that there is a calendar for every conceivable year?

Yes

No

Why?

There are $12$ choices for the months, and $2$ choices for the year $($leap or non$-$leap$).$ The total number of calendars is $12+2=14.$

There are $12$ choices for the months, and $2$ choices for the year $($leap or non$-$leap$).$ The total number of calendars is $12*2=24.$

There are $7$ choices for the day of January $1^{st}$ and $2$ choices for the year $($leap or non$-$leap$).$ The total number of calendars is $7+2=9.$

There are $7$ choices for the day of January $1^{st}$ and $2$ choices for the year $($leap or non$-$leap$).$ The total number of calendars is $7*2=14.$

There are $365$ choices for the day in a regular year and $366$ choices for the day in a leap year. The total number of calendars is $731.$