Quote:
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!
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
mvictor wrote:
excuse me, but WTF is P? where did u see such notations in official gmat questions??
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
MeghaP wrote:
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.
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 onsee 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 _________________