GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
S = {2, 4, 6} T = {2, 4, 6, 8, 10, 12} A new set M is to be construc
[#permalink]
Updated on: 02 Mar 2025, 14:23
Updated on: 02 Mar 2025, 14:23
Expert Reply
00:00
Question Stats:
0% (00:00) correct
100% (00:56) wrong based on 1 sessions
Hide Show timer Statistics
S = {2, 4, 6}
T = {2, 4, 6, 8, 10, 12}
A new set M is to be constructed such that, set S is a subset of M and set M is a subset of T. How many sets satisfying this property can be constructed?
A. 5
B. 6
C. 7
D. 8
E. 9
T = {2, 4, 6, 8, 10, 12}
A new set M is to be constructed such that, set S is a subset of M and set M is a subset of T. How many sets satisfying this property can be constructed?
A. 5
B. 6
C. 7
D. 8
E. 9
- Part of the project: GRE Quant & Verbal ADVANCED Daily Challenge 2025 NEW Edition - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
- Also replying to the unanswered questions
Re: S = {2, 4, 6} T = {2, 4, 6, 8, 10, 12} A new set M is to be construc
[#permalink]
Updated on: 02 Mar 2025, 14:20
Updated on: 02 Mar 2025, 14:20
Expert Reply
We have set $\(S=\{2,4,6\}\)$ and set $\(T=\{2,4,6,8,10,12\}\)$; we need to check the number of different sets of set $M$ can be formed if set $\(M\)$ is a subset of $\(T\)$ and $\(S\)$ is a subset of set $\(M\)$.
As we know set $\(M\)$ is a subset of set $\(T\)$ so set $M$ can have one more of the elements of set $\(T\)$. Also as we know that set S is a subset of set M , set M must have all its elements i.e. 2, 4 and 6 .
So, set M, other than the elements of set S i.e. 2,4 and 6 , can have none or more of $\(\{8,10,12\}\)$ i.e. none, one or more of the three elements is to be selected for set M .
Thus, the number of set M which can be formed with the given conditions is $\({ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3=1+3+3+1=8\)$
$\(\left(\because{ }^n C_r=\frac{n!}{r!\times(n-r)!}\right.\)$ is the number of ways to select $\(r\)$ things out of $\(\left.n\right)\)$
Hence the answer is (C).
As we know set $\(M\)$ is a subset of set $\(T\)$ so set $M$ can have one more of the elements of set $\(T\)$. Also as we know that set S is a subset of set M , set M must have all its elements i.e. 2, 4 and 6 .
So, set M, other than the elements of set S i.e. 2,4 and 6 , can have none or more of $\(\{8,10,12\}\)$ i.e. none, one or more of the three elements is to be selected for set M .
Thus, the number of set M which can be formed with the given conditions is $\({ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3=1+3+3+1=8\)$
$\(\left(\because{ }^n C_r=\frac{n!}{r!\times(n-r)!}\right.\)$ is the number of ways to select $\(r\)$ things out of $\(\left.n\right)\)$
Hence the answer is (C).
Re: S = {2, 4, 6} T = {2, 4, 6, 8, 10, 12} A new set M is to be construc
[#permalink]
26 Feb 2025, 10:42
26 Feb 2025, 10:42
Expert Reply
For me the official explanation is wrong ...........it says 9 but should be 7
$$
\(\begin{aligned}
& n(S)=3 \\
& n(m)=x \\
& n(T)=6 . \\
& S \subseteq m \subset T \\
& n(S)=2 \\
& M=\{2,4,6, \ldots\} \\
& T=S U\{8,10,12\} \\
& S S T .
\end{aligned}\)
$$
(1) $\(s=M\)$
(2)
$$
\(\begin{aligned}
\theta= & \{5,8\} \\
& \{5,10\} \\
& \{5,12\} \\
& \{5,8,10\},\{5,8,12\},\{5,10,12\} .
\end{aligned}\)
$$
$$
\(n(m)=1+3+3\)
$$
$$
\(=7\)
$$
$$
\(\begin{aligned}
& n(S)=3 \\
& n(m)=x \\
& n(T)=6 . \\
& S \subseteq m \subset T \\
& n(S)=2 \\
& M=\{2,4,6, \ldots\} \\
& T=S U\{8,10,12\} \\
& S S T .
\end{aligned}\)
$$
(1) $\(s=M\)$
(2)
$$
\(\begin{aligned}
\theta= & \{5,8\} \\
& \{5,10\} \\
& \{5,12\} \\
& \{5,8,10\},\{5,8,12\},\{5,10,12\} .
\end{aligned}\)
$$
$$
\(n(m)=1+3+3\)
$$
$$
\(=7\)
$$
