Permutation

hundreds digit is either 8 or 9.
(for 8 )
odd numbers end in 1,3,5,7,9. so 5 possibilities
8 possibilites left for the 2nd digit (can't be 8 or the units digit)
so 5*8=40 different odd numbers starting with 8
(for 9)
the same applies, but the units digit can't be 9, so it's 4*8 = 32

32+40=72 answer C.
is that the OA?

gviren wrote:How many odd thre digit numbers greater than 800 are there such that all their digits are different?

a) 40
b) 56
c) 72
d) 81
e) 104

OA after few replies

Answer is D 81

From 801 to 819 =9 numbers satisfy the cond
From 821 to 839 =9 numbers satisfy the cond
From 841 to 859 =9 numbers satisfy the cond
From 861 to 879 =9 numbers satisfy the cond
From 881 to 899 =9 numbers satisfy the cond

= 9*5 =45

From 901 to 919 =7 numbers satisfy the cond
From 921 to 939 =7 numbers satisfy the cond
From 941 to 959 =7 numbers satisfy the cond
From 961 to 979 =7 numbers satisfy the cond
From 981 to 999 =7 numbers satisfy the cond

= 7*5 = 35

45+35 =80

I might have missed one......

answer should be 81

jaspreet.sharma, actually 881 doesn't count because you can't use the same digit, so
from 881-899 (891, 893, 895, 897) only 4 numbers.(881, 883, 885, 887, 889, 899 can't be used)
9*4+4 = 36+4 = 40
from 981-999 (981, 983, 985, 987) only 4 numbers.(989, 991, 993,995,997 and 999 can't be used)
7*4+4 = 28+4 = 32

so 72

Jaspreet you r wrong..

881 to 899 is not 9 .. nothing from 880 to 889 will be counted due to duplicate 8..

AGPS - Fantastic approach..

Initially i was more towards figure of 81 but then cudnt find any error in ur approach... so i actually wrote down all numbers and striked off the incorrect ones..

BTG forum rocks !!!!

is there any way we can use the formula for permutation combinations to solve this one?.

I'm actually using it, just not calling it permutations....

"odd numbers end in 1,3,5,7,9. so 5 possibilities" this is 5P1 = 5!/4! = 5
"8 possibilites left for the 2nd digit (can't be 8 or the units digit)" this is 8P1 = 8!/7! = 8
"(for 9)
the same applies, but the units digit can't be 9, so it's 4*8 = 32 " instead of 5P1 you have 4P1

If the answer is 40, then my explanation goes like this:-

Since the question states that all three digits should be distinct and all the numbers selected should be odd, we can thus divide the question in two parts

1. Numbers 801 to 899

_____ _____ _____ (These represent the three digits)
The first digit has to be 8 therefore - 1
The third digit has to be odd so it can be (1,3,5,7,9) so - 5
The second digit then has to be (0,2,4,6) so - 4
(So the total numbers can be 5x4x1 = 20

2. Numbers 900 to 999

The first digit has to be 9 so - 1
The third digit can be (1,3,5,7) (here 9 cant be repeated) so - 4
The second digit can be (0,2,4,6,8) so - 5
(So the total numbers - 4x5x1 = 20

Adding 1 & 2 we get total numbers as 40

I guess the answer is 72

Num starting with 8 and odd =
1(1 For 8)*8 (Remaining) *5 (Any of {1,3,5,7,9}) = 40

Num starting with 9 and odd=

1(1 For 9)*8 (Remaining) *4 (Any of {1,3,5,7}) = 32


Total=40+32 = 72

vinaysingh wrote:If the answer is 40, then my explanation goes like this:-

Since the question states that all three digits should be distinct and all the numbers selected should be odd, we can thus divide the question in two parts

1. Numbers 801 to 899

_____ _____ _____ (These represent the three digits)
The first digit has to be 8 therefore - 1
The third digit has to be odd so it can be (1,3,5,7,9) so - 5
The second digit then has to be (0,2,4,6) so - 4
(So the total numbers can be 5x4x1 = 20

2. Numbers 900 to 999

The first digit has to be 9 so - 1
The third digit can be (1,3,5,7) (here 9 cant be repeated) so - 4
The second digit can be (0,2,4,6,8) so - 5
(So the total numbers - 4x5x1 = 20

Adding 1 & 2 we get total numbers as 40

I got A also. My explanation is roughly the same, but here was my thought process:

1) > 800 so hundreds digit is 8 or 9
2) odd so units digit is 1, 3, 5, 7, or 9
3) all digits different so the remaining digits that are not used should be in the tens digit
3) many restrictions so use most restrictive first (1, then 2, then 3)

Use 1 and 2 first then based on 3 figure out what the middle digit can be:

8xx: 8 x [1,3,5,7,9] --> the middle [] can be only be [0, 2, 4, 6 ]

9xx: 9 x [1,3,5,7] --> 9 cannot be repeated in units digit but 8 can now be used so middle [] is [0, 2, 4, 6, 8]

multiplying these you get:

8xx: (1) (4) (5) = 20
9xx: (1) (5) (5) = 20

so 40 total

Hi Mayonnai,
The Question asks How many odd three digit numbers greater than 800 are there such that all their digits are different?

i.e 800 -999

According to the explanation given by you & Vinay we have:

8xx: 8 x [1,3,5,7,9] --> the middle [] can be only be [0, 2, 4, 6 ]

if we apply this logic we can say:

first digit is 8, say we take 3rd digit as 1, then according to your logic

i.e 2nd digit should be [0, 2, 4, 6 ]

we can get nos such as

801 , 821, 841, 861

these are correct

what about these nos 831 , 851, 871 ,891
(all these are 3 digit ODD & with all three digits different)

dont they need to be counted as well

You got 40 coz you poeple did the following

8xx: 8 x [1,3,5,7,9] --> the middle [] can be only be [0, 2, 4, 6 ]

now here the last digit can be picked in 5 ways as you stated, but
what does this 5 mean

It can be interpreted as " there are 5 possible digits that can be used as the last digit of the required nos but at a time only one will be used as there is only one position of the last digit"

Your approch assumes that all 5 are used simultaneouly to fill the last position of one nos so we cannot use these 5 as middle hence you shrink the options of the middle nos ie. instead of 8 to just 4. This is incorrect.

The correct approach would be the one used by agps

i.e for 8XX
first digit can be selected in 1 way

last digit in 5 ways [1,3,5,7,9] but not 5 at a time (you have 5 options choose any one say for e.g. 1,)

so we have 8 possibiltes for the middle digit all excluding 8 & 1

so nos ways is 8 X 5 =40

& similarly for 9XX

and correct ams would be 72 only

Actually, as I was writing out my answer, I realized there was a flaw in my answer - which is the flaw that you mentioned. But I was about 90% through writing out the answer so finished the explanation anyways. But I realized I was excluding the possibility of odd digits in the tens digit when I should not have been. I was just too lazy to redo the math =P.

Thanks for pointing it out though.