M22-16

Question

How many three-digit integers greater than 710 exist such that all of their digits are distinct?

A. 198
B. 202
C. 207
D. 209
E. 212

Bunuel · Accepted Answer

Official Solution:

How many three-digit integers greater than 710 exist such that all of their digits are distinct?

A. 198
B. 202
C. 207
D. 209
E. 212

First, determine how many integers between 700 and 999 have all their digits distinct.
 
We have: 
 
 	ext{(3 options for the first digit)}*	ext{(9 options for the second digit)}*	ext{(8 options for the third digit)} =  
 
 = 216  numbers.
 
Among these 216 numbers, 9 of them (701, 702, 703, 704, 705, 706, 708, 709, 710) have distinct digits but are not greater than 710. Therefore, the answer to the question is  216 - 9 = 207 .

Answer: C

Siddhant26930 · Answer

Can we do this question as follows?

(1*8*8+1*9*8+1*9*8)-1. Basically taking each set of numbers separately and then subtracting 1 i.e. 710

I remember this kind of an approach from a question I did previously

Regards,

ekanth · Answer

i think you have missed one number 707

216-10

chetan2u · Answer

Hi,
216 is the total number where all digits are different..
707 has 7 at two places, so 707 has not been taken as a part of these 216 numbers..

happyface101 · Answer

Hi guys, I understand the solution above but have trouble seeing what I did wrong - can you please help?

If the first digit is 7: 2nd slot can be anything from #2 to #9 with the exception of 7, so that's 9-2+1-1=7; 3rd slot can be anything from #1 to #9 with the exception of #7 and the digit in the 2nd slot, so that's 9-1+1-2=7; so the total number of digit combinations is 7*7 = 49 numbers

If the first digit is 8: 2nd slot can be anything from #0 to #9 with the exception of 9, so that's 9-0+1-1=9; 3rd slot can be anything from #0 to #9 with the exception of #8 and the digit in the 2nd slot, so that's 9-0+1-2=8; so the total number of digit combinations is 9*8 = 72 numbers

If the first digit is 9: the combination is the same as if the first digit is 8. So, 72.

Together there are 49 + 2*72 = 193 combinations ...

What did I do wrong? Thank you!!

shubham1985 · Answer

Bunuel,

I solved the question in the below way. Please correct me if wrong.

Per the question, we have 3 options(7,8,9) for the first digit.

1)  1  8  8  = 64 ----- 7 is the first digit here

2)  1  9  8  = 72 ------ 8 is the first digit here

3)  1  9  8  = 72 ------- 9 is the first digit here

So total is 64+72+72 = 208.
However, in the first case we have to remove one number to account for 710. So answer is 207.

jasanisanket24 · Answer

I also approached the question in the above way.

Bunuel, can you please explain?

septwibowo · Answer

Try to explain :

- You did the correct calculation for 8 and 9 hundred digit.
- Anyway, you miscalculated for the 7 hundred digit :  don't forget the 0 number in the ten and unit digit . That's why, the number of digits possible is 1*8*7, not 1*7*7. After this,  you must add manually the number from 710-719  (which is 7 different number).

Hope it helps.

mrosale2 · Answer

Bunuel or All,

Why is 3*9*8 used to determine the amount of numbers that are from 700-799 and with three different integers?

Bunuel · Answer

(3 options for the first digit): 7, 8, or 9.

(9 options for the second digit): all digits except the one digit we used above.

(8 options for the third digit): all digits except the two digits we used above.

Check  Constructing Numbers, Codes and Passwords Questions  in our  Special Questions Directory .

Hope it helps.

houston1980 · Answer

Great explanation in the original solution!!! There are usually many ways to solve a problem.

Bunuel · Answer

Yes, that is also a correct approach.

Manku · Answer

Bunuel

I approached this question in the following way:

710-799

1 choice for 1st digit : 7
7 choices for 2nd digit : 2,3 5,6, 8,9,0
8 choices for 3rd digit : (10- 2)

total = 1*7*8= 56

800-899
1 choice for 1st digit : 8
9 choices for 2nd digit : 10-1 =9 
8 choices for 3rd digit : (10- 2=8)

total = 1*9*8= 72

900-999
1 choice for 1st digit : 8
9 choices for 2nd digit : 10-1 =9 
8 choices for 3rd digit : (10- 2=8)

total = 1*9*8= 72

Total 72*2 + 56=  200

I seem to have missed few of them ?
Please help

Thanks

Bunuel · Answer

You are missing numbers between 710 and 720: 712, 713, 714, 715, 716, 718, 719.

RK007 · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

varad98 · Answer

From 700-799:
We have one possibility at hundreds place(7), 9 possibilities at tens place (0,1,2,3,4,5,6,,8,9)& 8 possibilities for units place.
=1x9x8=72

From 800-899:
We have one possibility at hundreds place(8), 9 possibilities at tens place (0,1,2,3,4,5,6,7,,9)& 8 possibilities for units place.
=1x9x8=72

From 900-999:
We have one possibility at hundreds place (9), 9 possibilities at tens place (0,1,2,3,4,5,6,7,8,)& 8 possibilities for units place.
=1x9x8=72

There some 9 numbers which don't have repeating digits between 700-710 (inclu)
701,702,703,704,705,706,708,709,710

Total= (3x72)-9= 207

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

tanuja91 · Answer

So I read this as a combination question, vs a permutation one since the question asks for how many integers such that each digit is unique, i.e. 798 = 789 so we would have to exclude these overlaps...it would have helped if the question prompt included an example of what would meet the criteria to make sure the question is clear.

M22-16

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