
A box contains 6 balls which may be all of different colors or three each of two colors or two each of three different colors. The number of ways of selecting 3 balls from the box (if a ball of the same color is identical) is N, then find the value of ‘N’.
Answer
159.6k+ views
Hint: First we find out the number of ways the distribution can be done in all the 3 cases. The cases being 6 distinct colored balls or 6 balls three each of two colors or 6 balls of two each of three colors. We find the number of ways each case can be done and add them to find the final answer.
Complete step-by-step solution
A box contains 6 balls which may be all of different colors or three each of two colors or two each of three different colors.
So, the number of cases of possible options is 3.
Case 1: 6 distinct colored balls
We need to choose 3 balls out of 6 distinct ones which will be done in ${}^{6}{{C}_{3}}=\dfrac{6!}{3!\times 3!}=20$.
Case 2: 6 balls three each of two colors
We need to choose 3 balls. The choices can be 2 same-colored balls and 1 different or all three of same-colored balls.
The first part (2 white 1 black or 2 black 1 white) can be done in
$2\left[ {}^{3}{{C}_{2}}\times {}^{3}{{C}_{1}} \right]=2\times \dfrac{3!}{2!\times 1!}\times \dfrac{3!}{2!\times 1!}=18$.
The second part (3 white or 3 black) can be done in ${}^{3}{{C}_{3}}+{}^{3}{{C}_{3}}=1+1=2$.
Case 3: 6 balls of two each of three colors
We need to choose 3 balls. The choices can be 2 same-colored balls and 1 different or all three of different colored balls.
The first part can be done in choosing which color we are taking 2 balls off.
\[{}^{3}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}=3\times 2\times 2=12\].
The second part (1 white and 1 black and 1 red) can be done in \[{}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}={{2}^{3}}=8\].
Total number of options are $20+18+2+12+8=60$. The value of N is 20.
Note: We need to separately find the ways of choosing and we can use a table and actual color names to solve the problem. The multiplication of choosing is happening as the events are independent of each other.
Complete step-by-step solution
A box contains 6 balls which may be all of different colors or three each of two colors or two each of three different colors.
So, the number of cases of possible options is 3.
Case 1: 6 distinct colored balls
We need to choose 3 balls out of 6 distinct ones which will be done in ${}^{6}{{C}_{3}}=\dfrac{6!}{3!\times 3!}=20$.
Case 2: 6 balls three each of two colors
We need to choose 3 balls. The choices can be 2 same-colored balls and 1 different or all three of same-colored balls.
The first part (2 white 1 black or 2 black 1 white) can be done in
$2\left[ {}^{3}{{C}_{2}}\times {}^{3}{{C}_{1}} \right]=2\times \dfrac{3!}{2!\times 1!}\times \dfrac{3!}{2!\times 1!}=18$.
The second part (3 white or 3 black) can be done in ${}^{3}{{C}_{3}}+{}^{3}{{C}_{3}}=1+1=2$.
Case 3: 6 balls of two each of three colors
We need to choose 3 balls. The choices can be 2 same-colored balls and 1 different or all three of different colored balls.
The first part can be done in choosing which color we are taking 2 balls off.
\[{}^{3}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}=3\times 2\times 2=12\].
The second part (1 white and 1 black and 1 red) can be done in \[{}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{2}{{C}_{1}}={{2}^{3}}=8\].
Total number of options are $20+18+2+12+8=60$. The value of N is 20.
Note: We need to separately find the ways of choosing and we can use a table and actual color names to solve the problem. The multiplication of choosing is happening as the events are independent of each other.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
