Question:
Given \(5\) different green toys, \(4\) different blue toys, and \(3\) different red toys, how many combinations of toys can be chosen taking at least one green and one blue toy?
Given \(5\) different green toys, \(4\) different blue toys, and \(3\) different red toys, how many combinations of toys can be chosen taking at least one green and one blue toy?
Show Hint
If at least one object must be selected from \(n\) distinct objects, the number of selections is
\[
2^n-1
\]
because the empty set is excluded.
Updated On: Jun 24, 2026
- \(32\times 16\times 4\)
- \(31\times 15\times 4\)
- \(32\times 16\times 8\)
- \(31\times 15\times 8\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Select green toys.
There are \(5\) different green toys.
The number of ways to choose any subset of these toys is \[ 2^5=32 \] But at least one green toy must be selected, so exclude the empty selection: \[ 32-1=31 \] Thus, the number of ways to choose green toys is \[ 31 \]
Step 2: Select blue toys.
There are \(4\) different blue toys.
Total subsets: \[ 2^4=16 \] Since at least one blue toy must be selected, exclude the empty set: \[ 16-1=15 \] Thus, the number of ways to choose blue toys is \[ 15 \]
Step 3: Select red toys.
There are \(3\) different red toys.
Red toys may or may not be selected.
Hence, total possible selections are \[ 2^3=8 \]
Step 4: Apply multiplication principle.
Total number of combinations: \[ 31\times 15\times 8 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{31\times 15\times 8} \]
There are \(5\) different green toys.
The number of ways to choose any subset of these toys is \[ 2^5=32 \] But at least one green toy must be selected, so exclude the empty selection: \[ 32-1=31 \] Thus, the number of ways to choose green toys is \[ 31 \]
Step 2: Select blue toys.
There are \(4\) different blue toys.
Total subsets: \[ 2^4=16 \] Since at least one blue toy must be selected, exclude the empty set: \[ 16-1=15 \] Thus, the number of ways to choose blue toys is \[ 15 \]
Step 3: Select red toys.
There are \(3\) different red toys.
Red toys may or may not be selected.
Hence, total possible selections are \[ 2^3=8 \]
Step 4: Apply multiplication principle.
Total number of combinations: \[ 31\times 15\times 8 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{31\times 15\times 8} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Combinations Questions
- The number of six-digit natural numbers that can be formed with the digits 2, 3, 4, 0, 5, 6, 7, 8 is:
- The number of ways in which a committee of 7 members can be formed from 6 teachers, 5 fathers and 4 students in such a way that at least one from each group is included and teachers form the majority among them, is:
- If $^{n}C_{12} = ^{n}C_{8}$, then $^{n}C_{17} =$
- A student is to answer \(10\) out of \(13\) questions in an examination such that he must choose at least \(4\) from the first \(5\) questions. The number of choices available to him is:
- The least value of \(n\) so that \[ {}^{\,n-1}C_3+{}^{\,n-1}C_4\gt {}^{\,n}C_3 \] is
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions