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

1 2 3 4 5

Bunuel

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Bunuel

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Bunuel

GMAT Club Official Explanation:

Each person has n options, integers from 1 to n, inclusive. So the total number of possible outcomes (denominator) is n^n in all cases.

1. All choose different numbers:

The number of ways to assign n different numbers (1, 2, ..., n) to n people is n!

So probability = n!/n^n

2. All choose the same number:

This can happen in exactly n ways: all choose 1, all choose 2, ..., all choose n

So probability = n/n^n

We are given:

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

Cancel denominator:

n! = 6n

Substituting the answer choices, we see that only n = 4 satisfies the equation: 4! = 6 *4 = 24.

Answer: B.

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P_1 (n-different) :
1st person has n choices.
2nd person has n-1 choices.
3rd person has n-2 choices and so on ...
So P_1 = n!/n^n

P_2 (n-same) :
So can be done in n ways.
So P_2 = n/n^n

Given,
P_1 = 6*P_2
n! = 6n
(n-1)!=6
Thus,
n=4

Answer is B.

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total people given are n
and they select integer 1 to n inclusive
total options to choose n^n
all different numbers P is n!/(n^n)
P for same number is n/(n^n)
1/(n^n-1)

given that probability that all of them choose different numbers is six times the probability that all of them choose the same number

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

n^n * n! = 6*n
(n-1)! = 6
possible when n = 4

OPTION B , 4 is correct

Bunuel

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The permutation for n people to each pick a different integer is n!/n^n and that they all pick the same is n/n^n, or 1/n^n-1. Trying the answer choices, starting with the middle value first where n=5:

C) n=5
5!/5^5 = 24/625 is NOT 6 times larger than 5/5^5 = 1/625. Since this option is already to large, I'll try smaller values next.

B) n=4
4!/4^4 = 6/256 IS 6 times larger than 4/4^4= 1/256.

Answer: B)

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Bunuel

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Probability of different numbers is 1, probability of same number is 1/6 then we can take that n=6

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p(D) - Probability that all choose different numbers = n! ∕ (n^n)
p(S) - Probability that all choose the same number = n∕(n^n)
p(D) = 6 * p(S)
-> n! ∕ (n^n) = 6 * (n∕(n^n))
-> n! = 6n
-> (n-1)! = 6
Since 3! = 6, we have n-1 =3 , so n = 4

Answer : B

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Ways if n people choosing any number from 1 to n =n^n
Probability of all choosing same number = n/(n^n)
Probability of all choosing different number = n!/(n^n)
P(different number)=6*P(same number)
n!/(n^n)=6*n(n^n)
n!=6n
n(n-1)! =6n
(n-1)!=6
If n=2, LHS is not equal to RHS
IF n=3, LHS id not equal to RHS
If n=4, LHS= (4-1)!=3!=3*2=6

n=4

B

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So we know that there are n people and they will select from 1 to n integer.
lets find out n people having to select n number so total selections are = n^n
1. No of ways to choose different numbers = n * (n-1)* (n-2)... = n!
2. No of ways to choose same number = n (it can be that all choose 1 or 2 or 3 that way we have n choices)

P(different) = n!/n^n
P(same) = n/n^n

Now its given P(different) = 6 * P(same)
=> n!/n^n = 6 * n/n^n
=> n! = 6 * n
=> n (n-1)! = 6 * n
=> (n-1)! = 3*2*1
=> (n-1)! = 3!
=> n-1 = 3
=> n = 4

Hence Ans B

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Case 1 = All members choose different numbers,

Possible combinations = n!
Total combination = n^n
Hence probability of case 1 = n!/(n^n) ...... P1

Case 2 = All members choose the same numbers,

Possible combinations = n!
Total combination = n^n
Hence probability of case 1 = n/(n^n) ....... P2

Given, P1=6*P2
=>n!/n^n=6n/n^n
=> n!=6n

The above equation holds true when, n=4

Hence correct answer is choice B.

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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 probability that all of them choose different numbers = n*(n-1)*.....*1 = n!

The probability that all of them choose the same number = n*1*1...*1 = n

n! = 6n = n*3*2*1
n = 4

IMO B

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Probability of selecting all distinct:

All distinct: n!
All possibilities, without restriction: n^n

Therefore, probability of selecting all distinct: (n!)/(n^n)

Probability of selecting all equal:

All equal: 1 (x1 = x2 = ... )
"Possibility of equals": n (all equals 1, all equals 2, ...)
All possibilities, without restriction: n^n

Therefore, probability of selecting all equal: (n)*((1)/(n^n))

From stem we have:

P(all distinct) = 6 * P(all equal)

(n!)/(n^n) = 6 * (n)*((1)/(n^n))

(n!) = 6 * (n)*((1))

n! = 6 * n

(n!)/(n) = 6

(n-1)! = 6

n-1 = 3

Therefore, n = 4

Answer = B. 4

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Bunuel

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The probability of picking random number from 1...n, is $n!/n^n. $.

Similarly the probability of choosing one number for all is $nC1/n^n$.

Given that $n!/n^n. $ = 6* $nC1/n^n$

Here the best approach is to substute options, n! =6n. n=4 satisfies.

Hence IMO B

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Favourable outcomes for all of them to choose different numbers will be n!.[Because if there are 3 integers, the first person will have 3 options, the second person will have 2 options and last person will have 1option so it will 3!]

Favourable outcomes that all of them choose the same number will be n because there are n ways all of them can choose a number if n=3,then all of them can choose 1,2 or 3.So 3 favourable outcomes.

So n!=6n [No need to check total outcomes because it will cancel]
n=4

Hence answer is B.

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total cases= n*n*n...n times= n^n
different no.: n*n-1*n-2.....1 = n!
same no. = n*1*1*1....1=n

so, n!/n^n = n/n^n
(n-1)!=6
n=4

Ans B

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Total = n people choosing n different numbers = n^n

Number of ways where all of them choose the same number = n

Number of ways where all of them choose the different number = n!

P(all same)= n/n^n
P(all different) = n!/n^n

6*P(all same)= P(all different)
6*n/n^n= n!/n^n
6n=n!
6=n!/n
6=(n-1)!
n-1 = 3
n=4

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Firstly, the total number of combinations is
n x n

Distinct combinations
We begin by considering the first individual
The first individual has n choices
The second individual has (n-1) choices
....
The total number of distinct combinations is n!

Next, considering
All individuals are identical
This implies that if there are n numbers
There are only n combinations

For example
Three individuals
Three numbers
111
222
333
These are three combinations

Therefore, all identical implies n combinations

Given information in the problem
n! / n x n = 6n / n x n

(n-1)! = 6
(n-1)! = 3 x 2 x 1 ⇒ n = 4

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The total number of possible outcomes for n people each selecting an integer from 1 to n is n^n. The number of favorable outcomes for all people choosing different numbers is the number of permutations of n items, which is n!. The number of favorable outcomes for all people choosing the same number is n, as the group could choose any single number from 1 to n.

The probability of all choosing different numbers is n!/n^n, and the probability of all choosing the same number is n/n^n. The problem states that the first probability is six times the second, which establishes the equation n!/n^n = 6 * (n/n^n). Multiplying both sides by n^n cancels the denominators, leaving n! = 6n. Since n must be a positive integer, both sides can be divided by n, which simplifies the equation to (n-1)! = 6. The only integer for which the factorial is 6 is 3. Therefore, n-1 = 3, which means n = 4.

The value of n is 4. The correct answer is (B).

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Bunuel

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Number of ways each of n people randomly selects an integer from 1 to n = n!

Number of ways each of n people randomly selects the same integer = n

The denominator will be same and can be cancelled out.

n! = 6 * n

(n-1)! = 6

Therefore n - 1 = 3

n = 4

Option B