This problem involves the concept of combinations, which is a way of selecting items from a larger group where the order of selection does not matter. The formula for combinations is given by:

$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$

where $n$ is the total number of items available, and $k$ is the number of items to be selected. The exclamation mark (!) denotes the factorial, meaning the product of all positive integers up to that number.

We are given a batch of boys with different uniform colors:

• Red uniform boys: 4
• Green uniform boys: 7

The total number of boys in the batch is $4 + 7 = 11$.

We need to pick a squad of 5 boys from this batch.

Squad Selection: At Least One Red and One Green Uniform Boy (Part i)

For the first part, we need to determine the number of methods to pick a squad of 5 units such that the squad has at least one red uniform boy and at least one green uniform boy. This condition implies that the squad cannot consist solely of red boys or solely of green boys.

A common approach for "at least" problems in combinations is to use the complementary counting principle: find the total number of possible outcomes and subtract the number of undesired outcomes.

Total Squad Picking Methods

First, let's find the total number of ways to pick any 5 boys from the 11 available boys, without any restrictions. This is given by $\binom{11}{5}$.

$$ \binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!} $$

$$ \binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6!}{5 \times 4 \times 3 \times 2 \times 1 \times 6!} $$

$$ \binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} $$

$$ \binom{11}{5} = 11 \times 2 \times 3 \times 7 = 462 $$

So, there are 462 total ways to pick a squad of 5 boys from the batch.

Invalid Squads for Part (i)

Now, we identify the scenarios that violate the condition "at least one red and one green uniform boy":

• Squads with only red uniform boys:

  We have 4 red boys in total. It is not possible to pick a squad of 5 boys if all of them must be red, as we only have 4 red boys.

  $$ \binom{4}{5} = 0 $$

• Squads with only green uniform boys:

  We have 7 green boys. We need to pick 5 green boys from these 7.

  $$ \binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} $$

  $$ \binom{7}{5} = \frac{7 \times 6 \times 5!}{2 \times 1 \times 5!} = \frac{7 \times 6}{2} = 21 $$

  So, there are 21 ways to pick a squad consisting only of green uniform boys.

Calculation for Part (i)

The number of methods to pick a squad with at least one red and one green uniform boy is calculated by subtracting the invalid cases from the total ways:

Number of methods = Total ways - (Ways with only red boys + Ways with only green boys)

$$ 462 - (0 + 21) = 462 - 21 = 441 $$

Therefore, there are 441 methods to pick a squad with at least one red and one green uniform boy.

Squad Selection: At Least 3 Red Uniform Boys (Part ii)

For the second part, we need to figure out the number of methods to pick a squad of 5 units such that the squad has at least 3 red uniform boys. Since there are only 4 red boys available in the batch, this means the squad can have either 3 red boys or 4 red boys.

We will consider these two mutually exclusive cases and sum their possibilities.

Case 1: Exactly 3 Red Uniform Boys in the Squad

If the squad has exactly 3 red uniform boys, then the remaining $5 - 3 = 2$ boys must be green uniform boys to complete the squad of 5.

• Number of ways to choose 3 red boys from 4 red boys: $\binom{4}{3}$
• Number of ways to choose 2 green boys from 7 green boys: $\binom{7}{2}$

$$ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3!}{3! \times 1} = 4 $$

$$ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7 \times 6 \times 5!}{2 \times 1 \times 5!} = \frac{7 \times 6}{2} = 21 $$

The number of ways for Case 1 is the product of these combinations:

$$ \text{Ways (Case 1)} = \binom{4}{3} \times \binom{7}{2} = 4 \times 21 = 84 $$

Case 2: Exactly 4 Red Uniform Boys in the Squad

If the squad has exactly 4 red uniform boys, then the remaining $5 - 4 = 1$ boy must be a green uniform boy to complete the squad of 5.

• Number of ways to choose 4 red boys from 4 red boys: $\binom{4}{4}$
• Number of ways to choose 1 green boy from 7 green boys: $\binom{7}{1}$

$$ \binom{4}{4} = \frac{4!}{4!(4-4)!} = \frac{4!}{4!0!} = 1 \quad (\text{Note: } 0! = 1) $$

$$ \binom{7}{1} = \frac{7!}{1!(7-1)!} = \frac{7!}{1!6!} = \frac{7 \times 6!}{1 \times 6!} = 7 $$

The number of ways for Case 2 is the product of these combinations:

$$ \text{Ways (Case 2)} = \binom{4}{4} \times \binom{7}{1} = 1 \times 7 = 7 $$

Calculation for Part (ii)

The total number of methods to pick a squad with at least 3 red uniform boys is the sum of the ways from Case 1 and Case 2:

$$ \text{Total Ways (Part ii)} = \text{Ways (Case 1)} + \text{Ways (Case 2)} = 84 + 7 = 91 $$

Therefore, there are 91 methods to pick a squad with at least 3 red uniform boys.

Summary of Squad Picking Results

Based on our detailed calculations, the number of methods for (i) at least one red and one green uniform boy is 441, and for (ii) at least 3 red uniform boys is 91.