P & C

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 40
Joined: Sat Jan 30, 2010 10:38 am

P & C

by vijaynarayanan » Thu Jul 01, 2010 6:02 am
Question: A company plans to assign identification numbers to its employees. Each number is to consist of 4 different digits from 0 to 9, inclusive, except that the 1st digit cannot be 0. How many different identification numbers are possible?

I feel that the answer should be 9x10x10x10=9000.

But I see that the answer is 4536. Can someone explain how?

Thx
Vijay.

User avatar
Legendary Member
Posts: 1893
Joined: Sun May 30, 2010 11:48 pm
Thanked: 215 times
Followed by:7 members

by kvcpk » Thu Jul 01, 2010 6:27 am
vijaynarayanan wrote:Question: A company plans to assign identification numbers to its employees. Each number is to consist of 4 different digits from 0 to 9, inclusive, except that the 1st digit cannot be 0. How many different identification numbers are possible?

I feel that the answer should be 9x10x10x10=9000.

But I see that the answer is 4536. Can someone explain how?

Thx
Vijay.
Hi Vijay,

First place can be filled by 9 digits 1 to 9
second by 9 [because, we cannot repaet the first digit]
third by 8 [cant repeat 1 and 2 digits ]
fourth by 7

so total 9*9*8*7 = 4536

User avatar
Master | Next Rank: 500 Posts
Posts: 294
Joined: Wed May 05, 2010 4:01 am
Location: india
Thanked: 57 times

by amising6 » Thu Jul 01, 2010 6:58 am
vijaynarayanan wrote:Question: A company plans to assign identification numbers to its employees. Each number is to consist of 4 different digits from 0 to 9, inclusive, except that the 1st digit cannot be 0. How many different identification numbers are possible?

I feel that the answer should be 9x10x10x10=9000.
if you take

But I see that the answer is 4536. Can someone explain how?

Thx
Vijay.
good question
your reasoning makes sense because we are not talking about non repeatation
as for example 1111 and 2222 are all unique
Ideation without execution is delusion