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09 Jul 2025, 09:13
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Let n be the number of people and choices.
Total outcomes = n^n
Probability all choose different numbers:
P(different)= n!/ (n^n)
(This is the number of permutations divided by total outcomes).
Probability all choose the same number:
P(same)= n/(n^n) = 1 / {n^(n−1)}
(There are n ways for them to all choose the same number, e.g., all 1s, all 2s, etc.).
Given: P(different)=6×P(same)
n! / (n^n) = 6× 1/{n^(n−1)}
Multiply by n^n :
n!=6× {n^n} / {n^(n−1)}
n!=6×n^{n−(n−1)}
n!=6×n^1
n!=6n
Divide by n (since n is not equal to 0):
(n−1)!=6
We know that 3!=3×2×1=6.
So, n−1=3.
n=4
Answer: B
Total outcomes = n^n
Probability all choose different numbers:
P(different)= n!/ (n^n)
(This is the number of permutations divided by total outcomes).
Probability all choose the same number:
P(same)= n/(n^n) = 1 / {n^(n−1)}
(There are n ways for them to all choose the same number, e.g., all 1s, all 2s, etc.).
Given: P(different)=6×P(same)
n! / (n^n) = 6× 1/{n^(n−1)}
Multiply by n^n :
n!=6× {n^n} / {n^(n−1)}
n!=6×n^{n−(n−1)}
n!=6×n^1
n!=6n
Divide by n (since n is not equal to 0):
(n−1)!=6
We know that 3!=3×2×1=6.
So, n−1=3.
n=4
Answer: B
09 Jul 2025, 09:22
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NOTES:
n number of people
each selects a integer 1 to n
Probability of different numbers = (6)(probability of same number)
Find N
SOLVE:
Total outcomes: N^n
All the same number: N / N^n
All Different = N! / N^n
(6) ( N / N^n) = N!/N^n
Multiple N^n to remove that
N!= 6n
Divide by n
(n-1)!= 6
3!=6 so N =4
B: 4
n number of people
each selects a integer 1 to n
Probability of different numbers = (6)(probability of same number)
Find N
SOLVE:
Total outcomes: N^n
All the same number: N / N^n
All Different = N! / N^n
(6) ( N / N^n) = N!/N^n
Multiple N^n to remove that
N!= 6n
Divide by n
(n-1)!= 6
3!=6 so N =4
B: 4
