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04 Mar 2021, 07:15
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The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 are to be arranged around a circle. What is the probability that the even and odd numbers alternate ?
A. 1/9!
B. 1/126
C. 5/126
D. 1/12
E. 2/9
A. 1/9!
B. 1/126
C. 5/126
D. 1/12
E. 2/9
04 Mar 2021, 19:18
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There are 10 digits to be arranged. These have 5 even numbers.
Total Arrangements = (10 - 1)! = 9!
Arrangements where the even and odd numbers alternate
Arrangements for even numbers = (5 - 1)! = 4!
Arrangements for Odd numbers = 5! (as the even numbers serve as reference points)
Favorable Outcomes = (4! * 5!) = 24 * 5!
The required probability = (24 * 5!) / 9! = 24 / (9 * 8 * 7 * 6) = 1 / 126
Option B
Arun Kumar
Total Arrangements = (10 - 1)! = 9!
Arrangements where the even and odd numbers alternate
Arrangements for even numbers = (5 - 1)! = 4!
Arrangements for Odd numbers = 5! (as the even numbers serve as reference points)
Favorable Outcomes = (4! * 5!) = 24 * 5!
The required probability = (24 * 5!) / 9! = 24 / (9 * 8 * 7 * 6) = 1 / 126
Option B
Arun Kumar
27 May 2021, 03:04
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Since we have to find the probability of the possibility of alternative arrangement of even and odd numbers:
No. of even elements: 5
No. of odd elements: 5
Total No.: 10
probability: 5*5*4*4*3*3*2*2*1*1/10*9*8*7*6*5*4*3*2*1 or 5!*5!/10!=1/252
Since the arrangement of odd and even elements can be done in 2 ways
Total Probability= (1/252)*2=1/126
No. of even elements: 5
No. of odd elements: 5
Total No.: 10
probability: 5*5*4*4*3*3*2*2*1*1/10*9*8*7*6*5*4*3*2*1 or 5!*5!/10!=1/252
Since the arrangement of odd and even elements can be done in 2 ways
Total Probability= (1/252)*2=1/126
