Given $5$ different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is (A) $3600$ (B) $3720$ (C) $3800$ (D) $3600$

> [!TLDR] > We count all selections that include at least one green and one blue dye by first counting the total number of subsets from all 12 dyes, then subtracting those that violate the condition (no green or no blue). The answer is **3720**. The problem asks for the number of ways to choose any number of dyes (from 1 up to all 12) such that the selection contains **at least one green** and **at least one blue** dye. The red dyes are optional — they can be chosen or left out freely. This is a classic "at least one of each" counting problem. The key idea: instead of trying to list all valid combinations directly, we count the total number of subsets of the 12 dyes, then subtract the ones that are missing green or missing blue (or both). This is cleaner because "at least one" is easier to handle by complement. > [!FORMULA] > For a set of $n$ distinct items, the number of subsets (including the empty set) is $2^n$. > Here, each dye is distinct, so we treat each colour group separately. Let’s break it down. 1. **Total number of subsets of all 12 dyes** Since each dye can either be chosen or not, the total number of subsets (including the empty set) is $2^{12} = 4096$. 2. **Subsets with no green dye** If no green is chosen, we only pick from the 4 blue and 3 red dyes — that’s $4+3 = 7$ dyes. Number of subsets from these 7: $2^7 = 128$. This includes the empty set and all subsets that may or may not have blue or red. 3. **Subsets with no blue dye** If no blue is chosen, we pick from the 5 green and 3 red dyes — that’s $5+3 = 8$ dyes. Number of subsets: $2^8 = 256$. 4. **Subsets with neither green nor blue (only red)** These are counted twice in the above two steps, so we must add them back. Only red dyes: 3 dyes, so $2^3 = 8$ subsets (including the empty set). 5. **Apply inclusion-exclusion** Number of subsets that violate the condition (missing green OR missing blue) is: $128 + 256 - 8 = 376$. 6. **Valid su

Given 5 different green dyes, four different blue dyes and three different red dyes, the number of combination

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Concept understanding — Permutations Without Repetition

Permutations Without Repetition – The Idea of Arranging Things

Imagine you have three different books on a shelf: a Physics book, a Chemistry book, and a Maths book. How many different ways can you arrange them in a row?

You could try listing them out:

Physics, Chemistry, Maths
Physics, Maths, Chemistry
Chemistry, Physics, Maths
Chemistry, Maths, Physics
Maths, Physics, Chemistry
Maths, Chemistry, Physics

That's 6 arrangements. Notice that each arrangement uses all three books exactly once — no book is repeated, and no book is left out. This is the core idea: permutations without repetition count the number of ways to arrange a set of distinct objects in order, using each object exactly once.

Why "Without Repetition"?

The phrase "without repetition" means that once you place an object in a position, you cannot use it again. In our book example, once you put the Physics book in the first slot, you cannot put it in the second or third slot. Each object appears exactly once in the arrangement.

This is different from "permutations with repetition" (like creating 3-letter codes from the letters A, B, C where you can reuse letters — e.g., AAA, AAB, etc.). Here, no repeats allowed.

The Counting Logic – Why Multiply?

Let's build the arrangement step by step for 3 distinct books:

First position: You have 3 choices (any of the 3 books).
Second position: After placing the first book, only 2 books remain — so 2 choices.
Third position: Only 1 book is left — so 1 choice.

Total arrangements = $3 \times 2 \times 1 = 6$ .

This product $3 \times 2 \times 1$ is called 3 factorial, written as $3!$ .

P(n) = n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

For $n$ distinct objects, the number of permutations (arrangements in order) is $n!$ .

What If You Only Arrange Some of Them?

Suppose you have 5 different books, but you only want to arrange 3 of them on a shelf. How many ways?

First position: 5 choices
Second position: 4 choices
Third position: 3 choices

Total = $5 \times 4 \times 3 = 60$ .

This is a permutation of 5 objects taken 3 at a time, written as $P(5,3)$ or ${}^5P_3$ .

P(n, r) = \frac{n!}{(n-r)!} = n \times (n-1) \times \cdots \times (n-r+1)

Here $n$ is the total number of distinct objects, and $r$ is how many you are arranging. The formula works because:

Numerator $n!$ counts all arrangements of all $n$ objects.
Denominator $(n-r)!$ removes the arrangements of the $n-r$ objects you are not using.

Key Points to Remember

Important

Order matters — swapping two objects gives a different permutation.
No repetition — each object is used at most once.
For arranging all $n$ objects: $n!$
For arranging r out of $n$ objects: $\frac{n!}{(n-r)!}$

Common Mistake to Avoid …

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We count all selections that include at least one green and one blue dye by first counting the total number of subsets from all 12 dyes, then subtracting those that violate the condition (no green or no blue). The answer is 3720.

The problem asks for the number of ways to choose any number of dyes (from 1 up to all 12) such that the selection contains at least one green and at least one blue dye. The red dyes are optional — they can be chosen or left out freely.

This is a classic "at least one of each" counting problem. The key idea: instead of trying to list all valid combinations directly, we count the total number of subsets of the 12 dyes, then subtract the ones that are missing green or missing blue (or both). This is cleaner because "at least one" is easier to handle by complement.

n

Let’s break it down.

Total number of subsets of all 12 dyes

Since each dye can either be chosen or not, the total number of subsets (including the empty set) is $2^{12} = 4096$ .
Subsets with no green dye

If no green is chosen, we only pick from the 4 blue and 3 red dyes — that’s $4+3 = 7$ dyes.

Number of subsets from these 7: $2^7 = 128$ .

This includes the empty set and all subsets that may or may not have blue or red.
Subsets with no blue dye

If no blue is chosen, we pick from the 5 green and 3 red dyes — that’s $5+3 = 8$ dyes.

Number of subsets: $2^8 = 256$ .
Subsets with neither green nor blue (only red)

These are counted twice in the above two steps, so we must add them back.

Only red dyes: 3 dyes, so $2^3 = 8$ subsets (including the empty set).
Apply inclusion-exclusion

Number of subsets that violate the condition (missing green OR missing blue) is:

$128 + 256 - 8 = 376$ .
Valid subsets …

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Q.Given 555 different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is (A) 360036003600 (B) 372037203720 (C) 380038003800 (D) 360036003600

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Q.Given $5$ different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is
(A) $3600$
(B) $3720$
(C) $3800$
(D) $3600$

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