Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The number of bijective functions <img alt="\mathrm{f}:\{1,3,5,7, \ldots, 99\} \rightarrow\{2,4,6,8, \ldots, 100\}" src="/latex-image/?%5Cmathrm%7Bf%7D%3A%5C%7B1%2C3%2C5%2C7%2C%20%5Cldots%2C%2099%5

The number of bijective functions \mathrm{f}:\{1,3,5,7, \ldots, 99\} \rightarrow\{2,4,6,8, \ldots, 100\},such that \mathrm{\mathrm{f}(3) \geq f(9) \geq f(15) \geq f(21) \geq \ldots \geq f(99)}, is
 

Option: 1

{ }^{50} P_{17}
 


Option: 2

{ }^{50} P_{33}


Option: 3

33 ! \times 17 !


Option: 4

\frac{50 !}{2}


Answers (1)

best_answer

For bijective function

\mathrm{f(3)>f(9)>\ldots>f(99)}

First select 17 images for these 17 inputs \mathrm{^{50}C}_{33}  ways ad divide them in 1 way

Now selelct images for rest of the 33 inputs in 33! ways 

\mathrm{\therefore \text { Total }={ }^{50} \mathrm{C}_{17} \cdot 33 !}\\

                   =\frac{50 !}{17 !}\\

                   =50 P_{33}

Hence correct option is 2

Posted by

Kuldeep Maurya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Amrita BTech Admission 2025: JEE Main-based registration deadline extended to August 30

Latest Question