Question

Suppose a 5-digit PIN must be formed using the digits 0 through 4.
(b) How many such PINs contain exactly two 0s and exactly two 1s?
(c) List every such PIN that contains exactly two 0s and exactly two 1s.
(d) How many such PINs contain at least three 0s?

Answer 1

What do you think the answer is?

Answer 2

For b, I know it's 6 ways: 1100, 0011, 1010, 1001, 0110, 0101...that it's 6....and it is supposed to be 90...but I don't understand where the 15 comes from.

Answer 3

for a)
0,0,1,1 how many unique arrangements are there for this

Answer 4

3125

Answer 5

no thers just 6 xD

Answer 6

Oh, sorry. I thought it was how many in all.

Answer 7

Yes. I know. 6: 1100, 0011, 1010, 1001, 0110, 0101

Answer 8

okay instead of writing it out

Answer 9

can u figure out why it would be 4!/(2!*2!)

Answer 10

no

Answer 11

okay well think about it like this, so lets suppose theyre all different a,b,c,d
there are 4! ways to arragene them

Answer 12

but if a is b
then there are 2! ways you can arrange a b within themself which is not going to make a difference
a b
or
b a is the same if b= a

Answer 13

so 4!/2!
now if u have c and d also with the same property c= d
then u have
4!/(2!2!)

Answer 14

we can come back to that in a bit if u are still confused now

Answer 15

with these 6 you are sure of

Answer 16

why 2!?

Answer 17

for those 6, it will leave one of the 5 spots open as u only use 4 spaces,

Answer 18

right...so i will use 2,3,4, and 5

Answer 19

so 6*5, and then another 3 digits can, only 3 because 2,3,4 are left u cannot use another 0 or 1
so u have
6*5*3

Answer 20

a total of 90 ways

Answer 21

do u want me to explain with picture?

Answer 22

why 6*5?

Answer 23

okay so u understood why
0,0,1,1 have 6 unique arragenments

Answer 24

yes?

Answer 25

yes

Answer 26

now we can chose which one of the 5 spaces we are leaving out