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10 Jul 2025, 05:30
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total outcomes=n*n*n...(n times)=n^n
all of them choose different numbers: is a permutation of n elements, n!
all of them choose the same number: as there are n numbers, there are n possibilities
6*n/n^n = n!/n^n
6n=n!
6=(n-1)!
3=n-1 -> n=4
Correct answer is B
all of them choose different numbers: is a permutation of n elements, n!
all of them choose the same number: as there are n numbers, there are n possibilities
6*n/n^n = n!/n^n
6n=n!
6=(n-1)!
3=n-1 -> n=4
Correct answer is B
adityaprateek15
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10 Jul 2025, 05:41
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P(all diff) = n!/n^n
P (all same) = nC1/n^n
Given, n!/n^n = 6*nC1/n^n
n! = 6*nc1
n(n-1)! = 6n
n = 4
Option B
P (all same) = nC1/n^n
Given, n!/n^n = 6*nC1/n^n
n! = 6*nc1
n(n-1)! = 6n
n = 4
Option B
Bunuel
10 Jul 2025, 06:12
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Each of n people chooses a number randomly from 1 to n
We know that,
Probability(all different)=6×Probability(all same)
Now,
Total number of possible outcomes when each person independently chooses a number from 1 to n = n^n
Probability of all choosing the same number,
=> P(all same) = n/n^n
Probability of all choosing a different number,
=> P(all different) = n!/n^n
We already know as above mentioned,
Probability(all different)=6×Probability(all same)
=> n!/n^n = 6* n/n^n
=> n! = 6n
=> (n-1)! = 6
This is only possible when n = 4
B. 4
We know that,
Probability(all different)=6×Probability(all same)
Now,
Total number of possible outcomes when each person independently chooses a number from 1 to n = n^n
Probability of all choosing the same number,
=> P(all same) = n/n^n
Probability of all choosing a different number,
=> P(all different) = n!/n^n
We already know as above mentioned,
Probability(all different)=6×Probability(all same)
=> n!/n^n = 6* n/n^n
=> n! = 6n
=> (n-1)! = 6
This is only possible when n = 4
B. 4
