CPS 341 HOMEWORK 6 Due at class time, in class, Tuesday, 10 April, 2018 NOTE: (1

Question

CPS 341 HOMEWORK 6 Due at class time, in class, Tuesday, 10 April, 2018 NOTE: (1) You must write your solutions properly. Solutions without any ex- planation will receive no credit. Exhaustive enumeration is not an acceptable counting technique. (2) If you leave your answers in forms such as C(10, 3) x 4!, that would be fine (3) In the word MOMMY, all the M's are identical. (4) Recall that "or" means "either or or both" (1) Consider distinct permutations of the word MAXIMAL (a) (4 pts) How many are there that begin with M? (b) (6 pts) How many are there in each of which the two M's are not next to each other? (2) (15 pts) This problem concerns a standard deck of 52 cards. A hand consists of 5 cards a) How many hands either have exactly 2 diamonds or have exactly2 spades? (b) How many hands contain only cards of two kinds where 3 cards are of one kind and the other 2 are of the other kind (for example, ace of spades, ace of diamonds, ace of clubs, ten of diamonds, and ten of spades)? (c) How many hands contain 4 cards of the same kind (for example, 4 aces and ten of diamonds)?

Explanation / Answer

Solution:

Note: The first question is done as per Chegg guidelines, please repost others.

1

a)

MAXIMAL has 7 slotes in it -------, in which at the begining we need to put M

this means

M------

THis means they will be 6!

and total number of ways they can be arranged is: [C(2,1)x6!]/( 2!x2!)

b)

To compute number of permutation in which two M are not together i.e next to each other in following way. compute total number of permutation and subtract permutation in which two M are together.

Total number of permutation: 7!/( 2!x2!)

Total number of permutation with two M together is 6!/( 2!)

number of permutation in which two M are not together is: 7!/( 2!x2!)- 6!/( 2!)=5*(6!/(2!x2!))

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