Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive (A) $\frac{186}{190}$ (B) $\frac{187}{190}$ (C) $\frac{188}{190}$ (D) $\frac{18}{{}^{20}C_3}$

> [!TLDR] > The key idea is to count the total number of ways to choose 3 numbers from 1 to 20, then subtract the number of ways they *are* consecutive. The probability that they are **not** consecutive is $\frac{187}{190}$, which corresponds to option **(B)**. ## Concept and Intuition When a problem asks for the probability that "three numbers are **not** consecutive," the most natural approach is to use the **complement rule**: \[ P(\text{not consecutive}) = 1 - P(\text{consecutive}) \] Why? Because counting the cases where numbers *are* consecutive is much simpler than directly counting all the ways they could *not* be consecutive. The total number of ways to choose any 3 numbers from 1 to 20 is straightforward: it's $\binom{20}{3}$. The tricky part is counting the "consecutive" cases — but that turns out to be a neat, small count. > [!WARNING] > A common mistake is to think that "consecutive" means any three numbers that follow each other in order, like (5,6,7), but not necessarily starting at 1. That's correct — but students sometimes forget that the smallest possible consecutive triple is (1,2,3) and the largest is (18,19,20). There are exactly 18 such triples, not 20. ## Step-by-step solution 1. **Total number of ways to choose 3 numbers from 1 to 20** This is a simple combination: \[ \text{Total outcomes} = \binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140 \] 2. **Count the number of ways the three numbers are consecutive** If three numbers are consecutive, they must be of the form $(k, k+1, k+2)$. The smallest possible $k$ is 1 (giving 1,2,3). The largest possible $k$ is 18 (giving 18,19,20). So $k$ can be any integer from 1 to 18. That gives exactly **18** consecutive triples. …

Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive (A) (186)/(190) (B)

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\frac{186}{190}

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Concept understanding — Combinations Probability

Combinations Probability: Counting When Order Doesn't Matter

Imagine you are picking a team of 3 students from a class of 10. You do not care who is chosen first, second, or third — you only care which 3 students end up on the team. That is a combination: a selection where order does not matter.

If you now ask for the chance that one particular set of 3 students (say your three best friends) is the one chosen, you are doing combinations probability: probability where the favourable and total outcomes are both counted using combinations.

The Core Intuition

When every possible selection is equally likely (like drawing names from a hat), the probability of an event is the familiar ratio:

P(E) = \frac{\text{Number of favourable selections}}{\text{Total number of possible selections}}

This is the same "favourable over total" idea from basic probability — the only new part is that we count combinations, not arrangements, because order is irrelevant.

Note

The key difference from permutations: the group $\{$ Alice, Bob, Charlie $\}$ is the same selection as $\{$ Charlie, Bob, Alice $\}$ . Swapping the order of chosen items does not create a new outcome.

The Precise Statement

Let $n$ be the total number of distinct objects and let $r$ be how many you choose (without replacement). The number of ways to choose $r$ objects from $n$ is:

\binom{n}{r} = \frac{n!}{r!\,(n-r)!}

read as " $n$ choose $r$ ". If a selection of $r$ objects is made at random and every combination is equally likely, then the probability of an event $E$ is:

P(E) = \frac{\text{number of combinations in } E}{\dbinom{n}{r}}

A Worked Example

Problem: A bag has 5 red marbles and 3 blue marbles. You draw 3 marbles at random (without looking). What is the probability of getting exactly 2 red marbles?

Step 1 — Total outcomes. Choosing 3 marbles from 8:

\binom{8}{3} = \frac{8!}{3!\,5!} = 56

Step 2 — Favourable outcomes. You need exactly 2 red (from 5) and 1 blue (from 3):

\binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30

Step 3 — Probability.

P(\text{exactly 2 red}) = \frac{30}{56} = \frac{15}{28}

Tip

For "exactly $k$ of one type", multiply (ways to choose $k$ from that type) by (ways to choose the rest from the others), then divide by the total number of combinations.

Combinations vs. Permutations

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The key idea is to count the total number of ways to choose 3 numbers from 1 to 20, then subtract the number of ways they are consecutive. The probability that they are not consecutive is $\frac{187}{190}$ , which corresponds to option (B).

Concept and Intuition

When a problem asks for the probability that "three numbers are not consecutive," the most natural approach is to use the complement rule:

P(\text{not consecutive}) = 1 - P(\text{consecutive})

Why? Because counting the cases where numbers are consecutive is much simpler than directly counting all the ways they could not be consecutive. The total number of ways to choose any 3 numbers from 1 to 20 is straightforward: it's $\binom{20}{3}$ . The tricky part is counting the "consecutive" cases — but that turns out to be a neat, small count.

Watch out

A common mistake is to think that "consecutive" means any three numbers that follow each other in order, like (5,6,7), but not necessarily starting at 1. That's correct — but students sometimes forget that the smallest possible consecutive triple is (1,2,3) and the largest is (18,19,20). There are exactly 18 such triples, not 20.

Step-by-step solution

Total number of ways to choose 3 numbers from 1 to 20 This is a simple combination:

\text{Total outcomes} = \binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140

Count the number of ways the three numbers are consecutive If three numbers are consecutive, they must be of the form $(k, k+1, k+2)$ . The smallest possible $k$ is 1 (giving 1,2,3). The largest possible $k$ is 18 (giving 18,19,20). So $k$ can be any integer from 1 to 18. That gives exactly 18 consecutive triples. …

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Q.Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive (A) 186190\frac{186}{190}190186​ (B) 187190\frac{187}{190}190187​ (C) 188190\frac{188}{190}190188​ (D) 1820C3\frac{18}{{}^{20}C_3}20C3​18​

Concept and Intuition

Step-by-step solution

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Q.Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive
(A) $\frac{186}{190}$
(B) $\frac{187}{190}$
(C) $\frac{188}{190}$
(D) $\frac{18}{{}^{20}C_3}$

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