AMU 2026 — Mathematics PYQ
AMU | Mathematics | 2026a set contains elements. The number of subsets of the set which contain at most elements is:
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a set contains (2n+1) elements. The number of subsets of the set which contain at most n elements is:
2n
2n+1
2n−1
22n
(Correct Answer)22n
Let a set S have a total number of elements denoted by N, where:
N=2n+1
The number of ways to select a subset containing exactly r elements from a total pool of N elements is given by the binomial coefficient (rN).
The problem requires us to find the total number of subsets that contain at most n elements. This means the subset can have 0,1,2,…, up to n elements. We can write this sum as:
Total Subsets=(02n+1)+(12n+1)+(22n+1)+⋯+(n2n+1)
Let this desired sum be denoted by the variable S:
S=r=0∑n(r2n+1)— (Equation 1)
We know from the binomial theorem that the sum of all possible binomial coefficients for a power of 2n+1 equals 22n+1:
r=0∑2n+1(r2n+1)=22n+1
Expanding this full series completely gives us:
[(02n+1)+⋯+(n2n+1)]+[(n+12n+1)+⋯+(2n+12n+1)]=22n+1— (Equation 2)
According to the fundamental identity of binomial coefficients, we know that:
(rN)=(N−rN)
Applying this symmetry rule to the second half of our expansion in Equation 2:
(n+12n+1)=((2n+1)−(n+1)2n+1)=(n2n+1)
(n+22n+1)=((2n+1)−(n+2)2n+1)=(n−12n+1)
…
(2n+12n+1)=((2n+1)−(2n+1)2n+1)=(02n+1)
This shows that the second group of brackets contains terms that are identical in value to the first group of brackets. Therefore, both sections sum up to our target value S:
S+S=22n+1
2S=22n+1
Isolate S by dividing both sides of the equation by 2:
S=2122n+1
S=2(2n+1)−1
S=22n
The correct option is (d) 22n.
Let a set S have a total number of elements denoted by N, where:
N=2n+1
The number of ways to select a subset containing exactly r elements from a total pool of N elements is given by the binomial coefficient (rN).
The problem requires us to find the total number of subsets that contain at most n elements. This means the subset can have 0,1,2,…, up to n elements. We can write this sum as:
Total Subsets=(02n+1)+(12n+1)+(22n+1)+⋯+(n2n+1)
Let this desired sum be denoted by the variable S:
S=r=0∑n(r2n+1)— (Equation 1)
We know from the binomial theorem that the sum of all possible binomial coefficients for a power of 2n+1 equals 22n+1:
r=0∑2n+1(r2n+1)=22n+1
Expanding this full series completely gives us:
[(02n+1)+⋯+(n2n+1)]+[(n+12n+1)+⋯+(2n+12n+1)]=22n+1— (Equation 2)
According to the fundamental identity of binomial coefficients, we know that:
(rN)=(N−rN)
Applying this symmetry rule to the second half of our expansion in Equation 2:
(n+12n+1)=((2n+1)−(n+1)2n+1)=(n2n+1)
(n+22n+1)=((2n+1)−(n+2)2n+1)=(n−12n+1)
…
(2n+12n+1)=((2n+1)−(2n+1)2n+1)=(02n+1)
This shows that the second group of brackets contains terms that are identical in value to the first group of brackets. Therefore, both sections sum up to our target value S:
S+S=22n+1
2S=22n+1
Isolate S by dividing both sides of the equation by 2:
S=2122n+1
S=2(2n+1)−1
S=22n
The correct option is (d) 22n.