In how many ways can 3-digit numbers be formed selecting 3 d

Question

In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

A. 5P3/(2!)
B. 4P3
C. 4^3 
D. 4P3+3C1*(3!/2!)
E. 60 × 3!

chetan2u · Accepted Answer

We can form 3-digit numbers in two ways. 1) All digits different : there are four different digits so number of ways 3 digits can be chosen... 4P3 or 4C3*3! 2) Two digits similar: These 3-digit numbers will have two 1s and a digit from rest three - 2, 3, and 4.... \frac{3!}{2!} (number of ways when two digits are same) *3C1 (3 different digits)... so ans is 4P3+\frac{3!}{2!}*3C1 ... D Hi, P means permutation and is COUSIN, or I should say REAL BROTHER of C, combinations.. only thing is P is very concerned about the order/ sequence, so generally turns out to be not only real brother BUT also a BIG BRO of C .. so while 5C2 means \frac{5!}{3!2!} , 5P2 means \frac{5!}{3!2!} * 2! = \frac{5!}{3!} , as these two selected can be arranged in 2! ways.. so when you select it is C, and when you arrange, it is P.. Hope it clears some air around the the F of WTF, P

Hi, say we take these as five different digits.. let these be- 1=a, 1=b, 2=c, 3=d, 4=e.. now we have to choose three digits/letters so these numbers could be:- abc=112...bac=112....acd=123...bcd=123 and so on see here you are taking acd & bcd and abc & bac as two different 3-digit numbers, but they are the same.. so to avoid REPETITIONS we take them as 4 different digits so we consider 5 digits - 1,1,2,3,4- given as 4 different digits - 1,2,3,4- to find numbers with all different digits.. example 123,124,234 etc and then we consider similar digits as two digits and make combinations with remaining digits example 112, 121, 131,113 and so on Hope it helps

tomirisk · Answer

My answer is 4P3. We should discard a double digit. We only have 4 distinctive numbers to choose from.

flyingbunny · Answer

can't discard duplicated digits, for instance, the number can be 121.

the right one should be 5P3-3(3P2+3P1)=33

answer is C.

tomirisk · Answer

you are right flyingbunny that we can't discard a double digit. But answer C=4^3=64
and your solution 5P3-3(3P2+3P1)= 34. how can it be C?

flyingbunny · Answer

ha, 5P3-3(3P2+3P1)=60-3*9=33
answer is 4. 4P3+3C1*(3!/2!)=33.

bipolarbear · Answer

Umm... you want to explain your solution for laypeople? 8-)

atish · Answer

For those of you who need a detailed explanation
Lets find the number of 3 digit numbers with only single 1.
i.e. 3 digit numbers from 1,2,3,4
i.e. 4P3 (since order is important)

Next we find numbers with two 1s in it.
the third digit can be chosen in 3C1 ways and the numbers can be arranged in !3 ways, but 2 numbers are the same hence we divide by !2 (Formula used - number of ways of arranging p things which have q things of one kind, r things of another kind and so on is !p/(!q+!r+...))
Finally we have 3C1*(!3/!2))

Add the above two to get the answer D.
Hope this helps

ashueureka · Answer

I've a doubt regarding the answer and explanation given.
The question does not mention any constrain on the repetition of digits, so we can have 4 distinct choices for each position of the 3-digited number and hence we have
4^3 = 64 such numbers.
Can any one explain why this can't be the answer?

TIA.

DavidArchuleta · Answer

It means you just have 1 digit 2, 1 digit 3, 1 digit 4 and 2 digits 1 to form a number.

leve · Answer

This problem is written poorly. Anyways it has to state that 1,1,2,3,4 is a set of numbers. So you you can use each element (number) for once. From this perspective solution is easy: There are two cases. Numbers with single 1 and numbers with double 1. Which is 4P3 + 3C1 * 3C2 (Answer D)

mvictor · Answer

excuse me, but WTF is P? where did u see such notations in official gmat questions??

Vinayprajapati · Answer

In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

A. 5P3/(2!)
B. 4P3
C. 4^3
D. 4P3+3C1*(3!/2!)
E. 60 × 3!

Had it asked for all the different nos instead of 1,1,2,3,4 the ans wuld be: 5P3/2! = 30
In this solution we hav also halved (divided by 2!) the numbers containing 2,3 and 4. 
The numbers containing 2,3,4 are 6.
Add 6/2 =3 back to 30 and we get the answer i.e. 30+3=33.
Hence D is the answer.

thanks

bethebest · Answer

I am having trouble understanding this concept. Can you please explain why we cannot consider these as 5 different digits? How does the repetition of a number have an impact? Really grateful for an explanation.

chetan2u · Answer

Hi,

say we take these as five different digits..
let these be-
1=a, 1=b, 2=c, 3=d, 4=e.. 
now we have to choose three digits/letters 
so these numbers could be:-
 abc=112
bac=112
acd=123
bcd=123 and so on

see here you are taking acd and bcd ; and abc - bac as two different 3-digit numbers, but they are the same..
so to avoid REPETITIONS we take them as 4 different digits

so we consider 5- digits given as 4 different digits to find numbers with all different digits..
example 123,124,234 etc

and then we consider similar digits as two digits and make combinations with remaining digits
example
112, 121, 131,113 and so on

Hope it helps

bethebest · Answer

It makes sense now. Extremely daft of me to not see that. Thank you so much, much appreciated..!!

shashankism · Answer

Given digits : 1,1,2,3,4
There can be 2 cases for forming 3 digit numbers.

Case 1 : When 1 is used only once and we need to form the 3-digit number from digits 1,2,3,4 (Nos. like 123,234, etc.)  
No. of such numbers = 4P3

Case 2 : When 1 is used twice and 3rd digit is taken from 2,3,4 (Nos. like 113,141, etc.)  
No. of such numbers = 3C1*3!/2! (3C1 for selecting one number out of 3 numbers 2,3,4; 3! means arranging the 3 numbers 1,1 and 3rd selected number and it is divided by 2! as there are 2 1's. Due to which many numbers will be similar.)

Total numbers : 4P3 + 3C1*3!/2!

Answer D

JeffTargetTestPrep · Answer

We will separate the 3-digit numbers into two types: Those that contain two 1s and those that contain at most one 1.

1) The numbers that contain two 1s:

These numbers will be of the form _11, 1_1, or 11_, where _ is a digit of 2, 3, or 4. So, there are a total of 9 such numbers.

2) The numbers that contain at most one 1:

These numbers will consist of 3 of the digits 1, 2, 3, and 4. Since the order is important, there are 4P3 such numbers.

Scanning the answer choices, we see that D is the correct answer.

Answer: D

purvibagmar · Answer

However, why is 5P3/ 2! not thr right answer? We are dividing by 2! for the repitition

madkilla07 · Answer

Could someone please explain why 5P3/2! is wrong.

In how many ways can 3-digit numbers be formed selecting 3 d

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