Question

Question number 9

player A ends Up match, where the match ut r is (b) How many possible outcomes for I up with 4 points and B ends up with 3 points? (c) Now assume that they are playing a best-of when either player has 4 points or when 7 games hav ti For example, if after 6 games the score is 4 to 2 in favor of A, then A wins th s fitst are there, such that the match lasts for 7 games and A wins by a score of 4 to 3e 8. (a) How many ways are there to split a dozen people into 3 teams, where o and they don't play a 7th game. How many possible outcomes for the individual ga one team has 2 people, and the other two teams have 5 people each? (b) How many ways are there to split a dozen people into 3 teams, where each team ha people? (a) How many paths are there from the point (0,0) to the point (110, 111) in th plane such that each step either consists of going one unit up or one unit to the right (b) How many paths are there from (0, 0) to (210, 211), where each step consists of going one unit up or one unit to the right, and the path has to go through (110, 111)? To fulfill the requirements for a certain degree, a student can choose to take any 7 ou of a list of 20 courses, with the constraint that at least 1 of the 7 courses must be statistics course. Suppose that 5 of the 20 courses are statistics courses. (a) How many choices are there for which 7 courses to take? (b) Explain intuitively why the answer to (a) is not (1 et A and B be sets with |A n, B m ) How many functions are there from A to R (ie funt 19

Answer 1

Solution :-

Question -9 :-

( a ). How many paths are there from (0,0) to (110, 111), in the plane such that each step consists of going one unit up or one unit to the right?

Now consider, From (0, 0) to given point (110, 111) you have to take 'b' as upward steps and 'a' rightward steps. Now you have to take total of "a + b" steps. So the from "a + b" steps, we have to take 'a' rightward steps.

Hence the total number of paths from (0, 0) to (a, b) is $\binom{a+b}{c}$

$\binom{a+b}{c}=\frac{(a+b)!}{a!*(a+b-a)!}$

$\binom{a+b}{c}=\frac{(a+b)!}{a!*b!}$

Then, Number of paths from (0, 0) to (110, 111) is

$\binom{110+111}{110}=\frac{(110+111)!}{110!*111!}$

$\binom{110+111}{110}=\frac{221!}{110!*111!}$

( b ). How many paths are there from (0,0) to (210, 211), where each step consists of going one unit up or one unit to the right, and the path has to go through (110, 111)?

Now consider, Number of paths from (0, 0) to (210, 211) so that the parts has to go through (110, 111).i.e, the number of paths from (0, 0) to (110, 111) is multiplied with the number of paths from  (110, 111) to (210, 211).

Number of paths from (0, 0) to (210, 211) = number of paths from (0, 0) to (110, 111) * number of paths from  (110, 111) to (210, 211).

Number of paths from (0, 0) to (110, 111) is $\binom{110+111}{110}=\frac{221!}{110!*111!}$ ( $\because$From part(a))

Number of paths from  (110, 111) to (210, 211) is

( (210-110, 211-111)=(100, 100) )

$\binom{100+100}{100}=\frac{(100+100)!}{100!*100!}$

$\binom{100+100}{100}=\frac{(200)!}{100!*100!}$

Hence, Number of paths from (0, 0) to (210, 211) = $\frac{221!}{110!*111!}*\frac{(200)!}{100!*100!}$ ​​​​​​​