8. In an examination of 9 papers, a candidate has to pass in more papers than the number of papers in
which he fails in order to get the success. The number of ways in which he can fail
(A) 128 (B) 256 (C) 255 (D) 9 × 8!

Question

SAGAR SINGH - IIT DELHI · Answer

Dear student,

The number of ways he can fail=2^8-1=255

Hence c) is the correct option.

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..

Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

8. In an examination of 9 papers, a candidate has to pass in more papers than the number of papers in which he fails in order to get the success. The number of ways in which he can fail (A) 128 (B) 256 (C) 255 (D) 9 × 8!

8. In an examination of 9 papers, a candidate has to pass in more papers than the number of papers in
which he fails in order to get the success. The number of ways in which he can fail
(A) 128 (B) 256 (C) 255 (D) 9 × 8!

1 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Magical Mathematics[Interesting Approach]

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship