If some or all of $n$ objects are taken at a time, the number of combinations is $2^{n} - 1$.

If some or all of n objects are taken at a time, t – Class 11 Mathematics

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Concept understanding — Geometric Progression

Geometric Progression: The Idea of Repeated Multiplication

Imagine you're folding a piece of paper in half. Start with thickness 1 unit. After one fold, thickness becomes 2. After two folds, thickness becomes 4. After three folds, thickness becomes 8. The sequence of thicknesses is:

1, 2, 4, 8, 16, ...

Notice the pattern: each term is obtained by multiplying the previous term by the same number (here, 2). That's the core intuition behind a geometric progression — you keep multiplying by a fixed number, step after step.

This is different from an arithmetic progression, where you keep adding a fixed number. Here, the growth is multiplicative, not additive. That's why geometric progressions grow (or shrink) much faster.

Precise Definition

A Geometric Progression (GP) is a sequence of numbers where the ratio of any term to its preceding term is constant. This constant is called the common ratio, denoted by $r$ .

If the first term is $a$ , then the sequence looks like:

a, \ ar, \ ar^2, \ ar^3, \ ar^4, \ \dots

Note

The common ratio $r$ can be any real number — positive, negative, or even a fraction. If $r$ is negative, the terms alternate in sign. If $0 < r < 1$ , the terms get smaller and smaller.

The $n$ -th Term

To find any term directly without listing all previous ones, use the formula:

T_n = a \cdot r^{\,n-1}

where $T_n$ is the $n$ -th term, $a$ is the first term, $r$ is the common ratio, and $n$ is the term number (starting from 1).

Example: For the paper-folding sequence, $a = 1$ , $r = 2$ . The 5th term is $1 \cdot 2^{5-1} = 2^4 = 16$ , which matches our list.

Sum of $n$ Terms

There are two cases, depending on whether $r = 1$ or not.

n

When $r = 1$ , every term is just $a$ , so the sum is simply $n \times a$ .

Why the formula works (intuition):

Let $S = a + ar + ar^2 + \dots + ar^{n-1}$ . Multiply both sides by $r$ : $rS = ar + ar^2 + \dots + ar^n$ . Subtract the first from the second: $rS - S = ar^n - a$ , so $S(r-1) = a(r^n - 1)$ , giving the formula above.

Sum of an Infinite GP

If the common ratio $r$ lies strictly between $-1$ and $1$ (i.e., $|r| < 1$ ), the terms get smaller and smaller, and the sum of all terms approaches a finite value:

S_\infty = \frac{a}{1 - r}, \quad \text{for } |r| < 1

Watch out

If $|r| \geq 1$ , the infinite sum does not exist (it diverges to infinity or oscillates without settling). Never apply the infinite sum formula when $|r| \geq 1$ .

Example: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ has $a=1$ , $r=\frac12$ , so $S_\infty = \frac{1}{1 - 1/2} = 2$ . This matches the intuition that repeatedly halving a unit length eventually fills exactly 2 units.

Quick Reference Table

Property	Formula	Condition
Common ratio	$r = \dfrac{T_{n+1}}{T_n}$	Always
$n$ -th term	$T_n = a r^{n-1}$	Always

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The number of ways to select at least one object from $n$ distinct objects is $2^n - 1$ , because each object can either be chosen or not, and we exclude the case where none are chosen.

Concept and intuition

The problem is about counting combinations when we take "some or all" of $n$ distinct objects — meaning we choose at least one object. The key insight is that for each object, we have exactly two choices: include it or exclude it. If we consider all possible choices (including the case where we take none), the total number of subsets of a set of $n$ objects is $2^n$ . But the question specifically asks for combinations where at least one object is taken, so we simply subtract the one case where no object is chosen.

This is a classic result in combinatorics: the number of non-empty subsets of an $n$ -element set is $2^n - 1$ .

Step-by-step reasoning

Understand what "some or all" means

The phrase "if some or all of $n$ objects are taken at a time" means we are selecting any non-empty subset of the $n$ distinct objects. The order of selection does not matter — it's a combination, not a permutation.
Count all possible selections (including empty)

For each of the $n$ objects, we decide: take it or leave it. That gives $2$ choices per object. By the multiplication principle, the total number of ways to make these decisions for all $n$ objects is:

2 \times 2 \times \cdots \times 2 = 2^n

This $2^n$ counts every subset of the $n$ objects — from the empty set (choose none) to the full set (choose all).

Exclude the empty selection …

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Q.If some or all of nnn objects are taken at a time, the number of combinations is 2n−12^{n} - 12n−1.

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Q.If some or all of $n$ objects are taken at a time, the number of combinations is $2^{n} - 1$ .

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