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OCR CHAP 5C q: 7

Each of the digits 1,1,2,3,3,4,6 is written on a separate card . THe seven cards are then laid out in a row to form a 7- digit number.

A) How many distinct 7 digit numbers are there?
I got this one , it was 1260

b)How many of these 7 digit numbers are even
the answer is 540 ....but how?:frown:

c) How many of these 7 digit numbers are divisible by 4
the answer is 300 ....but how?:frown:

d) How many of these 7 digit numbers start and end with the same digit
the answer is 120 ....but how?:frown:

THANK yOU

Not for the faint hearted is it
..........ITS TOMORROW ARGHHHH
Reply 1
3/7 of the cards are even, they need to be on the end. (3/7)1260 = 540

this time the last two digits need to be a multiple of 4 - either 16,16,24,32,32. there are 7C2 = 21 different combinations of the last two numbers, so (5/21)1260 = 300

again, 21 combinations, this time only 1-1 and 3-3 are possibilities, so (2/21)1260 = 120
Reply 2
THANKs a bunch . Saved from a mental breakdown there. everything else i get , its just those. (shakes fist)
who invented permutations anyway.
GD luck in all ur exams everyone . im off 2 do sum more stats ...joy
b)In order to create a 7 digit number that is even, the last digit must be even --> 3 ways to pick the last digit.Then the remaining 6 cards must be arranged. It would seem that there are 6! ways (720) to do that, but there are some repeated digits that would form the same number. We don't care about the order of the 1s or the 3s, so we must divide by 2! and 2! for the ways to arrange the repeated digits.3 x 6! / (2! 2!)= 3 x 720 / 4= 3 x 180= 540 ways
(edited 4 years ago)
c)In order for the number to be divisible by 4, the last two digits must be a multiple of 4. Certainly the last digit must be even, but then we have to be careful about picking the tens digit. Specifically:

If the last digit is 2, then the tens digit must be odd (e.g. 12 or 32 are divisible by 4, but 42 and 62 are not)

If the last digit is 4, then the tens digit must be even (e.g. 24 or 64 are divisible by 4, but 14 and 34 are not)

If the last digit is 6, then the tens digit must be odd (e.g. 16 or 36 are divisible by 4, but 46 and 26 are not)

So the ending patterns are:

12 or 32 --> rest of digits contain 1 duplicate --> 2 x 5! / 2! = 120 ways

24 or 64 --> rest of digits contain 2 duplicates --> 2 x 5! / (2! 2!) = 60 ways

16 or 36 --> rest of digits contain 1 duplicate --> 2 x 5! / 2! = 120 ways
so 120+60+120= 300 ways
Reply 5
Original post by tashaa_ch
c)In order for the number to be divisible by 4, the last two digits must be a multiple of 4. Certainly the last digit must be even, but then we have to be careful about picking the tens digit. Specifically:

If the last digit is 2, then the tens digit must be odd (e.g. 12 or 32 are divisible by 4, but 42 and 62 are not)

If the last digit is 4, then the tens digit must be even (e.g. 24 or 64 are divisible by 4, but 14 and 34 are not)

If the last digit is 6, then the tens digit must be odd (e.g. 16 or 36 are divisible by 4, but 46 and 26 are not)

So the ending patterns are:

12 or 32 --> rest of digits contain 1 duplicate --> 2 x 5! / 2! = 120 ways

24 or 64 --> rest of digits contain 2 duplicates --> 2 x 5! / (2! 2!) = 60 ways

16 or 36 --> rest of digits contain 1 duplicate --> 2 x 5! / 2! = 120 ways
so 120+60+120= 300 ways

This must be a world record - you've resurrected a thread from 14 years ago :smile: