In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers m

Exercise 14.2 · Q11

Q. In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game? [Hint: order of the numbers is not important.]

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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 probability of winning is the ratio of one favourable outcome to the total number of ways to choose 6 numbers from 20, which is $\frac{1}{\binom{20}{6}} = \frac{1}{38760}$ .

This is a classic combinations probability problem. The key idea: the lottery numbers are drawn without regard to order — your set of six numbers either matches the committee’s set exactly or it doesn’t. There’s no “partial credit” and no sequence to worry about. So the probability is simply:

\text{Probability} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}

Here, there is exactly one favourable outcome: the specific set of six numbers chosen by the committee. The total number of possible outcomes is the number of different sets of six numbers you could pick from 1 to 20.

Because order doesn’t matter, we use combinations, not permutations.

r

Let’s work through it step by step.

Identify $n$ and $r$ .

You are choosing 6 numbers from the set $\{1, 2, 3, \dots, 20\}$ . So $n = 20$ , $r = 6$ .
Compute the total number of possible 6-number combinations.

Using the combination formula:

\binom{20}{6} = \frac{20!}{6! \, (20-6)!} = \frac{20!}{6! \, 14!}

Instead of expanding all factorials, cancel the common $14!$ :

\binom{20}{6} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1}

Simplify step by step. First, cancel the $6$ in the denominator with $18$ in the numerator:

\frac{20 \times 19 \times \cancel{18}^3 \times 17 \times 16 \times 15}{\cancel{6} \times 5 \times 4 \times 3 \times 2 \times 1}

Cancel the $5$ with $15$ :

\frac{20 \times 19 \times 3 \times 17 \times 16 \times \cancel{15}^3}{\cancel{5} \times 4 \times 3 \times 2 \times 1}

Cancel the $4$ with $16$ :

\frac{20 \times 19 \times 3 \times 17 \times \cancel{16}^4 \times 3}{\cancel{4} \times 3 \times 2 \times 1}

Cancel one $3$ with the $3$ in denominator:

$\frac{20 \times 19 \times \cancel{3} \times 17 \times 4 \times 3}{\cancel{3} \times 2 \times 1}$ …

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