1. A digit is rotatable if it is a recognizable digit when rotated 180. You can

Question

1. A digit is rotatable if it is a recognizable digit when rotated 180. You can assume the 1. A digit is rotatable f it is a recognizable digit when rotated 180. assume the only rotatable digits are 0, 1, 6, 8, and 9. The following problem involves 5-digit passcodes where the first digit cannot be 0 and digits cannot be repeated. a. How many passcodes are there? b. How many passcodes contain only rotatable digits? c. How many passcodes start with a rotatable digit or end with a non-rotatable digit?

Explanation / Answer

given that the rotatable digits are 0,1,6,8,9

pass code contains 5 digit and

give conditon

first cannot be 0 and digits cannot be repeated

now

a) how many pass codes are their

excluding 0 1st place can take 9 digits

and 2nd place can take except the first digit remaining 9 digits (0 can be included)

3rd place with 8 digits

4th place with 7 digits

5th place with 6 digits

total passcodes is 9*9*8*7*6=27216

b)

pass codes contains only rotatable digits

1st place excluding 0 we can take 4 digits

2nd place including 0 can take 4 digits

3rd with 3 digits

4th with 2 digits

5th with 1 digit

hence total is 4*4*3*2*1=96

c)

start with rotatble digit or end with a non rotatable

we can do like this

first calculate start with rotatable + end with non rotatable -start with roatable and end with non rotatble

i.e

n(aUb)=n(a)+n(b)-n(a(intersection)b)

n(a)=4*9*8*7*6=12096

first can take only 4 remaing are same as (a ) part

n(b)=8*6*7*5*8=13440

(0 can take values only in mid part of the digits )

and we can calulate without zero it is start from last digit

we can take last 5

next one 8 (since zero and the last are excluded)

next 7

next 6

next 5

total is 5*8*7*6*5

n(a(intersection)b)=4*6*7*8*5=6720

since first and last digits are independent

first 4

last 5

2nd 8

3rd 7

4rth 6

hence total answer of c is 12096+13440-6720=18816

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