This item is taken from Cambridge International AS and A Level Mathematics (9709) Pure Mathematics 1 Paper 11 of May/June 2010.
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To deal with this item, let us summarize the details of the problem:

** (a) Syllabus area: **Series* *
* (b) Formula/Concept needed: *
** **> Binomial Expansion
* > rth term or specific term of a binomial expansion*
* > Product of two binomials*

*Recall the Binomial Expansion Theorem and the rth term of the Binomial Expansion.*

*For part (i),*
*It means that*
*To do that, we need to identify the given values:*
*Substituting the values of a, b, n and r,*
*Simplify each of the terms*

**Hence, the first three terms of the expansion is **

For part (ii),
It means that
Take note that we are finding the
Since we are only interested in finding the coefficient of x, then
From part (i), we already have the term with x and x^3.
Simplifying and combining like terms,
*Hence, the coefficient of x in the product of the two binomials is 240. *

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*Your comments and suggestions are welcome here. Write them in the comment box below.*

Thank you and God bless!

This item is taken from IGCSE Additional Mathematics (0606) Paper 22 of October/November 2010.
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To deal with this item, let us summarize the details of the problem:

** (a) Syllabus area: **Permutation and Combination* *
* (b) Formula/Concept needed: *
** **> Permutation or Combination
* > Fundamental Principle of Counting*
* > Even numbers*
* > No Repetition of Digits*

Before we answer the items, let us encircle the important clues/phrases.
It means that
The next phrase is
It means that
Now, let us answer (i)
Note that the number has four (4) digits
Since there are no restrictions (except for the no repetition), we are considering all possible numbers that could be formed using the given digits. The first digit then can have
After using one on the first digit, the second digit will have
The third digit will then have
And the last digit will have
Multiplying the options for each digit,
*Hence, there are 840 (4-digit) numbers that could be formed from the given options.*

Another way of solving this item is to use Permutation (nPr). This is possible since order is important. That is

For (ii)
To make sure that the number is less than 4000, we will start with the first digit. A number is less than 4000 if the first digit is a 1, 2, or 3. However, in the given options, there are only two (1 or 3).
The remaining could be any digit,

Multiplying the options,
*Hence, there are 240 (4-digit) numbers that could be formed that are less than 4000.*
For (iii)
Again, to make sure that it is less than 4000, we will start with the first digit.
Further, to make sure that it is even, the last digit must contain an even digit. In the options given, only 4 and 8 are even.

The second and third may now have any of the remaining digits.
Multiplying the options,
*Hence, there are 80 (4-digit) numbers that could be formed that are even and less than 4000.*
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*Your comments and suggestions are welcome here. Write them in the comment box below.*

Thank you and God bless!

This item is taken from Cambridge IGCSE Additional Mathematics (0606) Paper 11 of October/November 2015.
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To deal with this item, let us summarize the details of the problem:

** (a) Syllabus area: **Permutation and Combination* *
* (b) Formula/Concept needed: *
** **> Permutation or Combination
* > Addition of Probabilities*

Before we answer the items, let us encircle the important clues/phrases.

It means that:
Next is
It means that:
The first case is that there is only one (1) woman. It means that five (5) men are needed to complete the team of 6. The total number of possible selections are:
The first case is that there are two (2) women. It means that four (4) men are needed to complete the team of 6. The total number of possible selections are:
The first case is that there are three (3) women. It means that three (3) men are needed to complete the team of 6. The total number of possible selections are:
Adding the number of possible selections:
**Hence, there are 203 teams that could be formed with 6 members and at least one woman.**

Another method is to consider all possible number of teams with 6 members that can be formed out of the 10. Here are the possible teams that could be formed without restrictions.
Let us separate the teams formed into two.
Since we are only interested on the number of teams with at least one woman, we could deduct the number of teams without any woman from the total. That is
The total number of teams formed with 6 members (no restriction) is
The total number of teams formed with 6 members without a woman is
or

Subtracting them,
*Hence, the total number of teams formed with 6 members with at least one woman is 203.*

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*Your comments and suggestions are welcome here. Write them in the comment box below.*

Thank you and God bless!