0

Guest

Courses New

We have 21 identical balls which needs to be distributed among 3 boys A,B,C such that A always gets even number of balls.Number of possible ways of doing this is less than or equal to.

We have 21 identical balls which needs to be distributed among 3 boys A,B,C such that A always gets even number of balls.Number of possible ways of doing this is less than or equal to.

Grade:11

2 Answers

Yash Patil
33 Points
5 years ago
162 ways.
 
No of Balls With A = 2
Balls Left = 19
ways of sharing 19 identical balls with B and C= 18
 
this one case , so total cases = 9 , But its not stated that B or C can be or cant be with no balls .
Assuming the can have no balls , total cases = 9
 
So total ways = 9x18= 162
 
less than or equal to 162
Randeep Sunil Bhargande
26 Points
2 years ago
 A + B + C = 19
Case I: If A = 0 then number of ways = 20C1
 Case II: If A = 2 then number of ways = 18C1
 Case III: If A = 4 then number of ways = 16C1
 : : : : 
 : : : :
 Case X : If A = 18 then number of ways = 2
C1
 ` total number of ways
 = 20C1
 + 18C1
 + 16C1
 + ... + 2
C1
 = 110

Think You Can Provide A Better Answer ?

Other Related Questions on IIT JEE Entrance Exam

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

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

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

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

Click Here Know More

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

Click Here Know More