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GMAT Club Olympics 2025 (Day 8): Each of n people randomly selects an

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given n people randomly chose from numbers 1 to n inclusive
P(all chose different numbers)=n(n-1)(n-2)........2*1/(n^n)=fact(n)/n^n----->1.
reason: first person has n ways to choose a no., second has (n-1) ways to choose a number and so on.. and total no of ways in which n different persons can choose from n different numbers=n^n
P(all chose same no.)=(1/n)^n
reason: P(any one person can choose a number)=1/n
for n persons to choose the same no=(1/n)^n------>2.
from given condition equation1.=6*equation2.
fact(n)/n^n=1/n^n =>fact(n)=6 or n=3
A.

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number of ways they can choose = n^n

number of ways all choose same number = n
number of ways all choose different number = n!

(n!/n^n) = 6*(n/n^n)

solving we get n = 4

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The number of ways that all of them choose different numbers is: $n!$

The number of ways that all of them choose the same number is: $n$

Based on the question: n!=6*n, if we write this like $n!=n*3*2$, we can conclude that $n=4$.

The answer is B.

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Total number of integers = n
total number of people = n

Total ways of selecting different numbers = n!
Total number of ways in which all people select same number = nC1
So n! = 6*nC1 = 6 * n
=> n! = 6n
=> (n-1)! = 6
=> n-1 = 3
=> n = 4

Hence answer is B

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Question strategy: We need to find an integer n such that in the given scenarium, the prob of all people different numbers "P(d)" is 6x the prob that all choose same number "P(s)". As we don't have any constraints about n, an efficient approach for the question is a Backsolving.
*Backsolving tip: When answer choices are ordered, always start with B or D, because you make sure that your maximum tries will be 2 (e.g If B is too big, you try D. If it is big, its C, if its small, its E)

Backsolving:
D) n = 6 -> P(s) = 1 x 1/6ˆ5 | P(d) = 1 x 5/6 x 4/6 x 3/6 x 2/6 x 1/6 = 120 / 6ˆ5 -> P(d) = 120 x P(s) ->Not ok

B) n = 4 -> P(s) = 1 x 1/4ˆ3 | P(d) = 1 x 3/4 x 2/4 x 1/4 = 6/4ˆ3 -> P(d) = 6 x P(s) -> Ok

-------------------
Each of n people randomly selects an integer from 1 to n, inclusive. If the probability that all of them choose different numbers is six times the probability that all of them choose the same number, what is the value of n?

A. 3
B. 4
C. 5
D. 6
E. 8

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Total number of possible outcomes
Each person independently selects a number from 1 to n, so:
Total outcome = $n^n$

1. Probability that all choose the same number
For all to choose the same number, they must all pick 1, or all pick n.
There are exactly n such favorable outcomes.
P(all same)= n/$n^n$

2. Probability that all choose different numbers
To have all different numbers:
First person: n choices, Second person: n−1 choices Third: n−2, and Last person: 1 choice
So:
P(all different)=n!/$n^n$

As per equation:

n!/$n^n$=6* n/$n^n$

Solving for n, we will get n=4.

Option B

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Bunuel

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Option B is the answer

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Total outcomes=n^n
total outcomes when everyone choose the same num= n/n^n (let say P(X))
total outcomes when everyone choose a different num=n!/n^n (let say P(Y))
given , P(Y)=6P(X)
=>n!/n^n=6 n/n^n
=>n!=6n
=>(n-1)!=6
n-1=3
n=4
IMO:B

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Topic(s)- Probability, Factorials, Exponents
Strategy- Number Testing, Algebra
Variable(s)- Probability All Different = "P(d)"; Probability All Same = "P(s)"
P(d) = 6*P(s)

Determine Probabilities by Testing #s
If n = 3: Integer set is {1, 2, 3}

1. P(d) = (# favorable outcomes) / (total # of outcomes)
i) # Favorable outcomes
= (3 #s can be chosen 1st) * (3-1=2 #s can be chosen 2nd) * (2-1=1 # can be chosen 3rd) = 3*2*1 = 3!
= n!
ii) Total # of outcomes
(3 total #s)*(3 total #s)*(3 total #s) = 3^3
= n^n
iii) P(d) = (n!) / (n^n)

2. P(s)
i) # Favorable outcomes
= (Favorable outcome = all 1's) + (Favorable outcome = all 2s) + (Favorable outcome = all 3s) = 3
= n
ii) Total # of outcomes
= n^n
iii) P(s) = (n) / (n^n)

3. P(d) = 6*P(s)
(n!) / (n^n) = 6 * (n) / (n^n)
n! = 6n

4. Test #s
a) n = 3: (3*2*1) =? (6*3) [NO]
b) n = 4: (4*3*2*1 =? (6*4) [YES]

Answer: B

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Bunuel

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Solution:

1. The total number of outcomes where n people choose different numbers

1st person can choose a number in n ways
2nd person can choose a number in n-1 ways
.
.
.
.
nth person can choose a number in 1 way

Therefore, the total number of outcome where n people can choose different numbers is n!.

2. The total number of outcomes where n people choose same number

1st person can choose a number in n ways.
Remaining (n-1) person can choose the number in 1 way.

Therefore, the total number of outcome where n people can choose same numbers is n.

3. Total number of possible outcomes

1st person can choose a number in n ways
2nd person can choose a number in n ways
.
.
.
.
nth person can choose a number in n ways

Therefore, the total number of possible outcomes is n^n.

Now, as given in question

(n!/n^n) = 6 (n/n^n)

i.e. n! = 6n
i.e. n(n-1)! = 6n
i.e. (n-1)!=6

Hence, n=4.

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Each of n people randomly selects an integer from 1 to n, inclusive. If the probability that all of them choose different numbers is six times the probability that all of them choose the same number, what is the value of n?

The no. of ways in which they can choose different nos.: n!
The no. of ways in which they can choose the same nos.: n (The first person has the liberty to select any number, other just has to follow that)

n(n-1)! = 6 * n

(n-1)! = 3 * 2 * 1 = 3!;

n-1=3; n = 4

A. 3
B. 4
C. 5
D. 6
E. 8

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Correct Answer: Option B 4

Total Outcome= n^n

For all same numbers,
P= n/n^n

For all different numbers,
P= n!/n^n

If we combine both the cases we get,
6* n/n^n = n!/n^n
By solving we get,
6n = n!
Now we need to keep values for n and find value of n where both are equal.

1) n=1,
6*1= 1!
This is incorrect.

2) n=2
6*2= 2!
This is also incorrect.

3) n=3
6*3= 3!
This is incorrect.

4) n=4
6*4= 4!
This is correct.
Hence 4 is correct answer as it satisfy both the conditions.

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B. 4

For same numbers you have to multiply the chance of getting one specific number (1/n) up to (n-1) times, since the picking of the first numbers may be any number:
(1/n)^(n-1)

For different numbers,
(n-1)/n * (n-2)/n * ... * 1/n = ((n-1)/n)! = (n-1)!/n^(n-1)

Now you have to solve 6 * (1/n)^(n-1) = (n-1)!/n^(n-1)
6 * [1^(n-1)] * ~~[(1/n)^(n-1)]~~ = (n-1)! * ~~[(1/n)^(n-1)]~~
6 * 1 = (n-1)!
3*2*1 = (n-1)!
n=4

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Each of n people randomly selects an integer from 1 to n, inclusive. If the probability that all of them choose different numbers is six times the probability that all of them choose the same number, what is the value of n?

Probability of chosing different numbers = n!/n^n

Probability of chosing same numbers=n/n^n

P(different)=6*P(Same)
n!=6n

(n-1)!=6

solving n=4
Answer is B

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Same integer:

Each person has $\frac{1}{n}$ probability of choosing an integer.

Probability of first person choosing any integer is $1$. Probability of every other $(n-1)$ person choosing the same integer is $\frac{1}{n}$.

$\text{Probability of choosing same integer }=\frac{1}{n^{n-1}}$

Different integer:

No of ways $n$ integers can be selected by $n$ persons $= n^n$

No ways different integers can be selected $= n!$

$\text{Probability of choosing different integer }=\frac{n!}{n^n}$

$\frac{n!}{n^n}=6\frac{1}{n^{n-1}}$

$n!=6n$

$\text{(n-1)!=6; 3!=6}$

$n-1=3$

$n=4$

Answer: B

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Bunuel

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Each of n people randomly selects an integer from 1 to n, inclusive.
Total sample space = n people and n integers = $n^2$

probability that all of them choose different numbers
P(D) = $\frac{{1 * 2 * 3 * (n-1) * n} }{ n^2}$
P(D) = $\frac{n! }{ n^2}$

probability that all of them choose the same number
P(S) = $\frac{n }{ n^2}$
Note - First one has 'n' choices, but rest all have only 1 hence numerator is 'n'

If the probability that all of them choose different numbers is six times the probability that all of them choose the same number
=> P(D) = 6 * P(S)
=> $\frac{n! }{ n^2}$ = $\frac{n }{ n^2}$
=> n! = 6n
=> (n-1)! = 6
=> n - 1 = 3
=> n = 4

Option B

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Case 1: If each of the n people chooses a different integer from 1 to n, then:
The first person has n choices.
The second person has (n − 1) choices.
The third person has (n − 2) choices, and so on.
Total ways for all to choose different numbers = n!.

Case 2: All choose the same number
If everyone chooses the same number, then they must all pick the same integer from 1 to n.
There are exactly n possible choices for which number they all agree on.
Number of favourable ways=n.

Total possible ways overall = n^n (since each of the n people can pick any of n numbers).

Given Condition
P(all different)=6×P(all same)
n!/n^n = 6x(n/n^n)
n! = 6n

Checking options given, 4 is the correct choice. Option B

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In this we know that probablity is determined by factorial. If there are 6 integers then probablity of different selection will be n!
Now we will get an equation ie-n!=6n
So n-1! should be 6 we know that 3!is 6 so n will be 4.
Option B

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Total outcomes: n^n, since each of n people picks a number from 1 to n.
Favorable outcomes (all same):
Probability= n/n^n
Favorable outcomes (all different): Probability= n!/n^n

n!/n^n=6.n/n^n
n!=6n
n!/n=6
(n-1)!=6
n = 4.

Answer: n=4.
Option B.

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Ans: B (4)

Total outcome of n people choosing from n different numbers = n^n

P(same) = favourable outcome/total outcome
favourable outcome = all the people choose the same number = first person chooses one number from n, and the rest have only that option
first person = n,1,1,1,1,1.......... (n-1 times 1)
P(same) = n/(n^n)

P(diff) = favourable outcome/total outcome
favorable outcome = first chooses 1 out of n, then the next person chooses from the remaining (n-1) so on.
n * (n-1) * (n-2) ......... 1 = n!
P(diff) = n!/n^n

given P(diff) = 6* P(same)
n!/n^n = 6 * n/(n^n)
n! = 6 * n

Only 4 satisfies this [Answer]

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