Question

permutations and combinations

Answer 1

The Gladstone Harbour board decides to issue private boats using the Gladstone Harbour with 7 digit registration numbers. None of these can start with zero. The harbour board had only been able to get the digits 0, 1, 2, 3, 4, 5, 7, 9 for the boat registration numbers. No digit may be used more than once in each registration number.

Bruno, who was responsible for allocating the registration numbers to the boats, had the bright idea of using the number 9 upside down to make 6. However, the 6 and 9 cannot both occur in the same registration number.

The registration numbers are issued in ascending order. Thus the first registration number is 1023567. The next two are 1023576 and 1023579.

How many boats can have registration numbers staring with 1?

Answer 2

okay, so fortunately the first digit is always 1, so we don't have to worry about it

Answer 3

We have 7 remaining digits to fill in the 6 remaining numbers

Answer 4

Yes.

Answer 5

This is 7 choose 6.

Answer 6

Then we have to count how many contain a 9

Answer 7

Wait... it's not 7 choose 6.... I was wrong. Order matters, so it is 7 permute 6

Answer 8

how would u write that into ur calculater

Answer 9

nPr I believe

Answer 10

However, this is not the complete story

Answer 11

okay

Answer 12

Now we need to find out how many contain a 9.

If they contain a 9 then we can change it to a 6, thus we can count it twice

Answer 13

7 P 6 gave us the total if we ignore the 9 flips to 6 fact.

Answer 14

okay

Answer 15

The easiest way I can think to count how many have 9, is to count how many DONT have 9 and subtract that from the total.

Answer 16

The number which don't have nine would be 6 P 6

Answer 17

so it is 7P6 - 6P6

Answer 18

Yes, so that gives us how many we can make by flipping 9 to 6

Answer 19

If you add that to our previous total, that will give us the adjusted total... which I believe is the answer.

Answer 20

Well, that would give us our answer if we started at 1000000
We need to subtract out any between 1000000 and 1023567
However, I think 1023567 is the first one what matches the no repeating digits criterion

Answer 21

Wait no... 1023456 comes before it...

Answer 22

So there a few between 1000000 and 1023567 which we need to subtract out... I'll leave that to you.

Answer 23

here we have 9 digits right ? if we fix first digit as 1, we're left with 8 more digits (including flip + constrain 6 and 9 cannot exist simultaneously)

Answer 24

Cursory glance... 7! + 6x6! seems to work. I can be wrong...

Answer 25

okay so can you explain how you got 7! + 6x6 for my explanation... thanks alot, here is a cupcake XD

Answer 26

lol its like this :-
first assume 6 is not there. we oly have 8 digits to mess wid. since we fixed 1 in first space, we have 6 more spaces to fill, and 7 digits to pick from : 7x6x5x4x3x2 = 7!

Answer 27

next 6 strikes in and we have to give away 9.

so 6 spaces to fill, and 7 digits to play wid again : 0, 2, 3, 4, 5, 6, 7

you must fill 1 space wid 6, and exclude any one of the remaining 6 digits, 6 times :- 6! x 6

Answer 28

@ganeshie8 Doing it that way makes you double count cases which do not have 6 or 9.

Answer 29

can we use this formula: c=n!/(n-r)!

Answer 30

Doing it my way does not have that double counting.

Answer 31

@wio nopes. if you see the 6x6! part, there we're considering the new strings that have 6 oly. so there is no question of double counting. 7! + 6x6! looks correct - ive double checked :)

Answer 32

@ozhobbits we have been using that same formula all the way :- 7! = 7!/(7-6)!

Answer 33

k

Answer 34

@ozhobbits lay back... take a deep breath, go over the problem again... you wil understand im sure :) good luck !

Answer 35

k