probabilty questions:

Question

1.A PIN number for a bank card consists of 4-6 digits, with no restrictions. How many such PIN numbers are possible? (Repetition is allowed and 0 is allowed in first position) [Hint: consider 3 cases]  



2.How many permutations are there of the letters in the word BASKETBALL?

1. With no restrictions.                                                                                        
2. The two Ls must be together?                                                                         

3. How many different 7-digit telephone numbers can a given area code have if the first digit cannot be zero?               (Repetition is allowed)                                                                                                  

4.There are eight teams competing in the Toronto District Mathematics Challenge. In how many ways can you choose the first, second, and third place team?                                                  

5.A team of 7 students is to be selected from a class of 30.                               

1. a.  How many different teams can be selected?                                                                               
2. b.  Aaron, Brad and Charles are students in this class. What is the probability that the team includes these 3 students?                                                                                                              
3. c.  How many teams would not include all three students?                                                        

Answer 1

Pankhudi

(1). Obviously 3 cases - 4, 5 and 6 digits.

For the 4 digits each digit can be 0-9 (10 possibilities) so number of possible pins = 10 x 10 x 10 x 10 = 10000

Do the same calculation for the 5 and 6 digit case, and add up the number of ways

(2) As there are 10 letters if we ignored that some letters repeat (B,A,L) there would be 10 x 9 x x 1 = 10 ! permutations. As there are two B's we need to divide this by 2! , same for the A, L, so number of permutations is 10 ! / (2 ! 2! 2!)

If the LL have to be together ... as there are 10 places, there are 9 ways we can put LL together - 1st/2nd place, 2nd/3rd etc.

Once the LL is placed, there are 8 letters to place in the 8 vacant spots. with a doubled B and A. So we can do this in 8 ! / 2! 2! ways.

So final answer is 9 x ( 8! / 2! 2! )

(3) 1st digit 1-9 (9 ways). Other 6 digits 0-9 (10 ways) , so 9 x 10⁶

⁽⁴⁾ fFirst can be chosen in 8 ways, second in 7, third in 6. So 8 x 7 x 6

(5) (a) This is 30 choose 7 = 30 C 7 = 30 ! / (7! 23 !)

(b) How many ways can we choose a team with these three. Well if we put them in the team, we have to choose 4 more students from a possible 27. So 27 choose 4 = 27 ! / (23 ! 4!) ways. To get the probability divide this by the total ways ie your answer from (a)

(c) You can just take away the number of ways from (b) from the number of ways in (a) here

MIKE