Suppose a 4-digit PIN must be formed using the digits 0 through 9. (e) How many

Question

Suppose a 4-digit PIN must be formed using the digits 0 through 9.

(e) How many such PINs do not repeat any digit?

(f) How many such PINs do not repeat a digit consecutively? For example, 9129 does not repeat a digit consecutively, even though it repeats the digit 9. However, 3341 does repeat the digit 3 consecutively.

(g) How much such PINs have the property that the first two digits sum to 10?

(h) How much such PINs have the property that the first digit is an 8 and the last two digits sum to 8?

(i) How much such PINs have the property that both the first two digits and the last two digits sum to 10?

(j) List every such PIN that contains exactly two 0s and exactly two 1s. How many are there?

Explanation / Answer

e>
we need to form a 4 digit pin with no digits are to be repeated
total numbers = 10
so first digit could take 1 out of all the 10 numbers that is 10 ways
the second place will have to take 1 from 9 numbers that is 9 ways
likewise 8 ways for the third digit
and 7 ways for the last digit
hence the total number of such pins would be = 10*9*8*7 = 5040 pins

f>
the first place will have 10 ways
the second place cannot have the same digit as that in the first place so 9 ways
the third again will have 9 ways as it can take the number in the first place but not the number in the second place
likewise 9 ways for the fourth place
hence the total number of such pins would be = 10*9*9*9 = 7290

g>
the first two digits of the pin must add up to 10
and there is no restriction on the other two digits so the last two digits could have any of the 10 digits so 110 ways each
for the last two digits
now if the first digit is 1 then the second need to be 9 , these digits could be swapped amongst themselves as well , so 2 ways
hence with 1 and 9 we'll have a total of = 2*1*1*10*10 = 200 ways , we multiplied by 2 as the 1 and the 9 could be arranged
amongst themselves as well

likewise when we choose 2 and 8 as the first two digits
total ways = 2*1*1*10*10 = 200 ways

when we choose 3 and 7
total ways = 2*1*1*10*10 = 200 ways

when we choose 4 and 6
total ways = 2*1*1*10*10 = 200 ways

when we choose 5 and 5
total ways = 1*1*10*10 = 100 ways

hence the total number of such pins whose first two digits add upto 10 are = 200+200+200+200+100 = 900 pins

h>
we need to fix the first digit as 8 so this place will be filled in 1 way
the second digit will be filled in 10 ways
when the last 2 digits are 0 and 8
total ways = 1*10*2*1*1 =20

when the last two digits are 1 and 7
total ways = 1*10*2*1*1 = 20

when the last two digits are 2 and 6
total ways = 1*10*2*1*1 = 20

when the last two digits are 3 and 5
total ways = 1*10*2*1*1 = 20

when the last two digits are 4 and 4
total ways = 1*10*1*1*1 = 10

hence the total number of pins with the first digit as 8 and the last two digits adding up to 8 are
= 20+20+20+20+10 = 90 ways

i>
we'll get the sum as 10 using the combination of (1,9) , (2,8) , (3,7) , (4,6) and (5,5)
now if the first two digits are 1 and 9 then we could have any of the above 5 combinations for the last two digits
1 and 9 could be arranged in 2 ways
and the combination for the last two could be arranges in 9 ways , 2 ways for all wxcept for (5,5) it'll have 1 way
=> total ways= 2*9 = 18 pins

when the first two digits are (2,8)
total ways = 2*9 = 18 ways

when the first two digits are (3,7)
total ways = 2*9 = 18 ways

when the first two digits are (4,6)
total ways = 2*9 = 18 ways

when the first two digits are (5,5)
total ways = 1*9 = 9 ways

hence total such pins would be = 18+18+18+18+9 = 81 ways

j> pins with exactly 2 zeros = 10*10*1*1 = 100
pins with exactly 2 = 100

therefore total pins= 100+100 = 200

there are 200 such pins

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