Math, asked by Amlenvikram, 9 months ago

if in a hall there are 10 randomly selected students then how many numbers of ways are there such that of them have different birthday. assume that all of then have their birthday in non leap years?

Answers

Answered by nike18
0

Answer:

3628800

Step-by-step explanation:

10*9*8*7*6*5*4*3*2*1

Answered by amitnrw
0

Given :   in a hall there are 10 randomly selected students

To find : numbers of ways  such that all of them have different birthday.

Solution:

in a hall there are 10 randomly selected students  

all of then have their birthday in non leap years

=> Days in a years  = 365

numbers of ways are there such that all  of them have different birthday

=> we need to select 10 Birthdays out of 365   and those 10 birthdays can be arranged in 10! ways

= ³⁶⁵C₁₀ * 10 !   or ³⁶⁵P₁₀

=  365! / 355!

= (365)(364)(363)(362)(361)(360)(359)(358)(357)*356)

numbers of ways are there such that all  of them have different birthday = 365! / 355!

numbers of ways are there such that none of them have different birthday

=> 1 birthday has to be selected out of 365

Hence 365 Ways

numbers of ways are there such that none of them have different birthday = 365

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