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# Dealing cards, counting bagels, permutations of a string

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How many ways are there to deal hands of seven cards to each of five players from a standard deck of 52 cards?

A bagel show has onion, poppy seed, egg, salty, pumpernickel, sesame seed, raisin, and plain bagels. How many ways are there to choose:
a) six bagels?
b) a dozen bagels?
c) two dozen bagels?
d) a dozen bagels with at least one of each kind?
e) a dozen bagels with at least three egg bagels and no more than two salty bagels?

How many permutations of the letters ABCDEFG contain:
a) the string BCD?
b) the string CFGA?
c) the strings BA and GF?
d) the strings ABC and DE?
e) the strings ABC and CDE?
f) the strings CBA and BED?

##### Solution Summary

The expert examines dealing cards, counting bagels and permutation of a string.

##### Solution Preview

In cards hands order does not matter, hence we deal with combinations.
First we must choose 7 cards for the first player from a deck of 52, which makes this number:
(1.1)
For the next player we choose 7 cards out of 45 cards that are left in the deck:
(1.2)
For the third player we choose 7 out of 38 cards:
(1.3)
For the fourth player we choose 7 out of 31 cards:
(1.4)
And for the last player we choose 7 cards out of 24:
(1.5)
Together the total number combinations is the product of all the combinations for single players:

(1.6)
Note that this is the same as choosing 35 cards from 52 cards deck when order is important (permutations):
(1.7)
This accounts all the possible cards permutations for the players.
And then for each hand we calculate the number of permutations which are 7!
Since order is not important, we must divide (1.7) by the number of each player permutations dividing 5 times) so again we get:
(1.8)

We have 6 ...

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