
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
133.2k+ 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
Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

JEE Main 2023 (April 6th Shift 2) Maths Question Paper with Answer Key

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

Current Loop as Magnetic Dipole and Its Derivation for JEE

Inertial and Non-Inertial Frame of Reference - JEE Important Topic

The maximum number of equivalence relations on the-class-11-maths-JEE_Main

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

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
