Question

just xiii thanks

(xi) How many different 5-card poker hands can be dealt from a regular 52-card (xii) How many of these hands contain no aces? (xii) How many contain a aces, for a-0 to 4? (xiv) How many contain all cards of the same suit? deck?

Answer 1

xi)

Possible Poker Hands in 52-Card Deck. The total number of possible hands can be found by taking the combination between 52 and 5. That is the number of ways 5 cards can be chosen out of 52 cards.  This means that if there are 52 cards, how many combinations of 5 cards can be drawn in $\binom{52}{5}$ways.

That is,

$\binom{52}{5}= \frac{52!}{47!\times 5!} =2598960$

xii)

if there is no Ace, that means the cards have to be dealt from a deck where there is no ace. he total number of possible hands can be found by taking the combination between 48 and 5. That is the number of ways 5 cards can be chosen out of 48 cards.  This means that if there are 52 cards, how many combinations of 5 cards can be drawn in $\binom{48}{5}$ways.

That is,

$\binom{48}{5}= \frac{48!}{47!\times 5!} =1712304$

xiii)

The required answer is,

$\binom{52}{5}+\binom{51}{5}\times 4 + \binom{50}{5}\ times 6 + \binom{49}{5}\times 4 +\binom{48}{5}=34047600$$\binom{52}{5}+\binom{51}{5}\times 4 + \binom{50}{5}\times 6 + \binom{49}{5}\times 4 +\binom{48}{5}=34047600$

xiv)

Five cards of the same suit mean the cards will be from either of the four suits. The required number of ways is,

$\binom{13}{5}\times 4=1287$