First , you should clarify it. Do you mean a particular king will be adjacent to a particular ace or at least one king would be adjacent to an adjacent to at least one ace or a particular king would be adjacent to at least one ace or at least one king will be adjacent to a particular ace ?

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThe problem has been solved here

The required probability is: \(\displaystyle 1- \dfrac{44!\cdot \sum_{k=1}^{3} \binom{3}{k-1}(4!) \binom{45}{k}\, (45-k)\, (46-k)\, (47-k)\, (48-k)}{52!} = \dfrac{284622747}{585307450} \approx 0.486279043603494\)

Log in to reply

First , you should clarify it. Do you mean a particular king will be adjacent to a particular ace or at least one king would be adjacent to an adjacent to at least one ace or a particular king would be adjacent to at least one ace or at least one king will be adjacent to a particular ace ?

Log in to reply

Any king adjacent to any ace.

Log in to reply

Probably, any king would be adjacent to any ace.

Log in to reply

18 but jatin is wright

Log in to reply