As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

A. 9%
B. 12%
C. 16%
D. 20%
E. 25%

out of curiosity guys, sometimes there are questions that we will have to look for the opposite and then subtract that answer from 1. but how come it doesn't work with this problem? for example:

what is the probability that they will choose the same number instead of different numbers:

4/4 * 1/4 * 1/4 * 1/4 = 1/64

so 1 - 1/64 = 63/64 which is not even close to 9%. how come this doesn't work? cause i always confuse between when to use the opposite and when not. anybody?

Possible combinations of each choosing a diffrent number (favorable events):
4*3*2*1=24

Total possible combinations (total events): 4*4*4*4=256

Favorable events/total events = 24/256 approx 10%

Is this a correct approach?

very quickly: total area of probability equals 4*4*4*4=4^4
at the numerator: 4! (if the first choose one number, there are 3 possible alternatives remaining...)
so, 4!/4^4=0.093....

p for all of them select 1 = 1/4^4

p for all of them select 1 or 2 or 3 or 4 = 4*1/4^4= 1/4^3
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As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

a) 9%
b) 12%
c) 16%
d) 20%
e) 25%

Soln:
Chances that all four choose different numbers is
Assuming first person chooses any of 4 numbers, second can choose the 1 from the left over 3, third can choose one number from the left over two and the last person chooses the last number. Hence
4 * 3 * 2 * 1 = 24 ways

Total possible chances is 4 * 4 * 4 * 4 = 256

Thus probability is = (24/256) * 100 = 9%

SOLUTIONS FOR ALL SCENARIOS

When four people choose an integer between 1 and 4, inclusive 5 cases are possible:

A. All choose different numbers - {a,b,c,d};
B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c};
C. 2 people choose same number and other 2 also choose same number - {a,a,b,b};
D. 3 people choose same number - {a,a,a,b};
E. All choose same number - {a,a,a,a}.

Some notes before solving:
As only these 5 cases are possible then the sum of their individual probabilities must be 1: P(A)+P(B)+P(C)+P(D)+P(E)=1

Probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}

As each person has 4 options, integers from 1 to 4, inclusive, thus denominator, total # of outcomes would be 4^4 for all cases.

A. All choose different numbers - {a,b,c,d}:

P(A)=\frac{4!}{4^4}=\frac{24}{256}.

# of ways to "assign" four different objects (numbers 1, 2, 3, and 4) to 4 persons is 4!.

B. Exactly 2 people choose same number and other 2 choose different numbers - {a,a,b,c}:

P(B)=\frac{C^2_4*4*P^2_3}{4^4}=\frac{144}{256}.

C^2_4 - # of ways to choose which 2 persons will have the same number;
4 - # of ways to choose which number it will be;
P^2_3 - # of ways to choose 2 different numbers out of 3 left for 2 other persons when order matters;

C. 2 people choose same number and other 2 also choose same number - {a,a,b,b}:

P(C)=\frac{{C^2_4*\frac{4!}{2!2!}}}{4^4}=\frac{36}{256}.

C^2_4 - # of ways to choose which 2 numbers out of 4 will be used in {a,a,b,b};
\frac{4!}{2!2!} - # of ways to "assign" 4 objects out of which 2 a's and 2 b's are identical to 4 persons;

D. 3 people choose same number - {a,a,a,b}:

P(D)=\frac{C^3_4*4*3}{4^4}=\frac{48}{256}.

C^3_4 - # of ways to choose which 3 persons out of 4 will have same number;
4 - # of ways to choose which number it will be;
3 - options for 4th person.

E. All choose same number - {a,a,a,a}:

P(E)=\frac{4}{4^4}=\frac{4}{256}.

4 - options for the number which will be the same.

Checking: P(A)+P(B)+P(C)+P(D)+P(E)=\frac{24}{256}+\frac{144}{256}+\frac{36}{256}+\frac{48}{256}+\frac{4}{256}=1.

Hope it's clear.
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# of ways to divide group of 4 into two groups of 2 when order of the groups does not matter is \frac{C^2_4*C^2_2}{2!} and then you can do 4*3 for numbers;

But if you do just C^2_4*C^2_2 then you get the # of divisions of group of 4 into two groups of 2 when the order of the groups matters (you'll have group XY and also group YX with this fromula) then for numbers you should use C^2_4.

So your formula needs to be divided by 2! in any case to get rid of the duplications.

Check the following links for more on this issue:
combination-anthony-and-michael-sit-on-the-six-member-87081.html?hilit=dividing#p767453
probability-88685.html?hilit=factorial%20teams
ways-to-divide-99053.html?hilit=factorial%20teams
combinations-problems-95344.html?hilit=factorial%20teams

Hope it's clear.
_________________

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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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Thanks for the reply Bunuel !

Exactly 2 people choose the same number and other 2 choose different numbers - {a,a,b,c}: 4*4*3*2/4^4 what's wrong with it, plz explain?