4. Given a 5-card hard from a 52-card deck: (a) A sequence is a hand consisting of five consecutive cards of any suit (e. g., 5♥ − 6♥ − 7♠ − 8♦ − 9♣). An ace may be either high (as in 10-J-Q-K-A, or low, as in A-2-3-4-5, but can’t "wrap around" (Q-KA-2-3 is not a valid sequence). How many different sequence hands are possible?

# sequences = 10,200

Step-by-step explanation:

- If we order the 5-card hand from highest number to lowest, the first card

may be one of the following: ace, king, queen, jack, 10, 9, 8, 7, 6, or 5.

- There are 10  possibilities:

- After the first card, the numbers showing on the remaining  four cards are completely determined.

-  If we allow flushes, including royal  flushes, there are four possible suits for each of the cards. (Note that this  holds because the cards all show different numbers, and there are four  suits for each number.)

               # straights = 10·4C1·4C1·4C1·4C1·4C1 = 10·45 = 10,240

- The number of straight flushes we get:

    There are 9 possibilities.  After the first card, whose suit we may choose in 4 ways, the remaining  cards are completely determined.

               # straight flushes = 9 4C1 = 9·4 = 36

-  The number of royal flushes we get:

     If we order the 5-card hand from highest card to lowest, the first card will  be an ace. There are four possible suits for the ace. After that, the other

four cards are completely determined. Thus, there are 4 possible royal

flushes:

               # royal flushes = 4C1 = 4

- Subtracting the number of straight flushes and royal flushes. Hence, we have:

               # sequences = 10,240 - 36 - 4

               # sequences = 10,200

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !