11 and 12 10) How many 5-card poker hands are there with a exactly 3 kings? b. a

Question

11 and 12

10) How many 5-card poker hands are there with a exactly 3 kings? b. all red cards? c. at least one black card? d. at least one card from each suit? e. one pair (two cards of the same rank and three cards, all with distinct ranks)? 11) Twenty seven different books are placed into 9 distinguishable boxes. In how many ways can this be done if a. there are no restrictions on the number of books per box? each box has 3 books? six boxes have 1 book each and the other three have 7 each? b. c. 12) A pizza place has 10 toppings available for pizza. How many pizzas are possible if a. b. c. d. e. any number of toppings is allowed, no doubling? any number of toppings is allowed, including doubling (but not tripling, etc.)? the pizza has exactly 4 toppings, no doubling? the pizza has exactly 4 toppings, including doubling? the pizza has at least 3 toppings, no doubling?

Explanation / Answer

Answers to Q11

Part (a)

No restriction

The first book can go into any one of the 9 boxes in 9 ways.

Since there is no restriction, the second book also can go into any one of the 9 boxes in 9 ways.

And this argument is true for all the 27 books.

Thus, total number of possibilities is: 9 x 9 x …….. x 9 – 27 times

= 927 ANSWER

Part (b)

Each box has 3 books.

Any 3 out of 27 books can be put in first box in 27C3 = (27!)/{(3!)(24!)}

Now, only 24 books are available and hence, the second box can have 3 books in

24C3 = (24!)/{(3!)(21!)}.

Arguing identically, the third box can have 3 books in 21C3 = (21!)/{(3!)(18!)}

And so on,

The last box can have 3 books in 3C3 = (3!)/{(3!)(0!)} .

Thus, the total number of possibilities = (27C3)( 24C3)( 21C3)……( 3C3)

= (27!)/{(3!)9} ANSWER

Part (c)

1book each in 6 boxes and 7 books each in 3 boxes

The 6 boxes having 1 book each, can be any 6 out of 9 boxes in (9C6).

Once these 6 boxes are selected, each box can be given a book in

(27 x 26 x 25 x 24 x 23 x 22) [first box can have any one of 27 books, second box can have any one of the remaining 26 books, third box can have any one of the remaining 25 books, and so on sixth box can have any one of the remaining 22 books]

The remaining 21 books can be distributed 7 books each in 3 boxes in (21!)/{(7!)3}

Thus, the total number of possibilities = (9C6) (27 x 26 x 25 x 24 x 23 x 22) (21!)/{(7!)3}

= (27!)/{60(7!)2}ANSWER

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