Understanding the Invitation Combination Problem
This problem involves calculating the number of ways to select a group of guests based on specific criteria. We need to invite a total of 3 women and 3 men. The key condition is that exactly 3 of the invited guests must be relatives of the man, and exactly 3 must be relatives of his wife.
Available Relatives Breakdown:
Let's list the relatives available for invitation:
- Man's Relatives: \(4\) women and \(3\) men (Total \(7\))
- Wife's Relatives: \(3\) women and \(4\) men (Total \(7\))
Setting Up the Variables and Conditions
To solve this, we can define variables representing the number of people invited from each category:
- Let \(x\) = Number of women invited from the man's relatives.
- Let \(y\) = Number of men invited from the man's relatives.
- Let \(z\) = Number of women invited from the wife's relatives.
- Let \(w\) = Number of men invited from the wife's relatives.
The problem gives us the following conditions:
- The total number of people invited from the man's relatives must be 3: \(x + y = 3\).
- The total number of people invited from the wife's relatives must be 3: \(z + w = 3\).
- The total number of women invited must be 3: \(x + z = 3\).
- The total number of men invited must be 3: \(y + w = 3\).
We also have constraints based on the number of available relatives:
- \(0 \le x \le 4\) (Man's women)
- \(0 \le y \le 3\) (Man's men)
- \(0 \le z \le 3\) (Wife's women)
- \(0 \le w \le 4\) (Wife's men)
Solving the System of Equations
We can express \(y\), \(z\), and \(w\) in terms of \(x\):
- From \(x + y = 3\), we get \(y = 3 - x\).
- From \(x + z = 3\), we get \(z = 3 - x\).
- Substitute \(z\) into \(z + w = 3\): \((3 - x) + w = 3\), which simplifies to \(w = x\).
Now, let's check if these expressions satisfy the availability constraints. We already know \(0 \le x \le 4\). Let's check the constraints for \(y, z, w\):
- Constraint for \(y\): \(0 \le y \le 3 \implies 0 \le 3 - x \le 3\). This inequality holds true when \(0 \le x \le 3\).
- Constraint for \(z\): \(0 \le z \le 3 \implies 0 \le 3 - x \le 3\). This also holds true when \(0 \le x \le 3\).
- Constraint for \(w\): \(0 \le w \le 4 \implies 0 \le x \le 4\). This condition is satisfied since we found \(0 \le x \le 3\).
Therefore, the possible integer values for \(x\) are \(0, 1, 2,\) and \(3\). We need to consider each of these possibilities.
Calculating the Number of Ways for Each Case
We use the combination formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) to calculate the number of ways to choose \(k\) items from a set of \(n\) items.
Case 1: \(x = 0\)
If \(x=0\) women are chosen from the man's relatives:
- \(y = 3 - 0 = 3\) men from man's relatives.
- \(z = 3 - 0 = 3\) women from wife's relatives.
- \(w = 0\) men from wife's relatives.
Number of ways = (Ways to choose 0 women from 4) \(\times\) (Ways to choose 3 men from 3) \(\times\) (Ways to choose 3 women from 3) \(\times\) (Ways to choose 0 men from 4)
Ways\(_1 = \binom{4}{0} \times \binom{3}{3} \times \binom{3}{3} \times \binom{4}{0} = 1 \times 1 \times 1 \times 1 = 1\).
Case 2: \(x = 1\)
If \(x=1\) woman is chosen from the man's relatives:
- \(y = 3 - 1 = 2\) men from man's relatives.
- \(z = 3 - 1 = 2\) women from wife's relatives.
- \(w = 1\) man from wife's relatives.
Number of ways = (Ways to choose 1 woman from 4) \(\times\) (Ways to choose 2 men from 3) \(\times\) (Ways to choose 2 women from 3) \(\times\) (Ways to choose 1 man from 4)
Ways\(_2 = \binom{4}{1} \times \binom{3}{2} \times \binom{3}{2} \times \binom{4}{1} = 4 \times 3 \times 3 \times 4 = 144\).
Case 3: \(x = 2\)
If \(x=2\) women are chosen from the man's relatives:
- \(y = 3 - 2 = 1\) man from man's relatives.
- \(z = 3 - 2 = 1\) woman from wife's relatives.
- \(w = 2\) men from wife's relatives.
Number of ways = (Ways to choose 2 women from 4) \(\times\) (Ways to choose 1 man from 3) \(\times\) (Ways to choose 1 woman from 3) \(\times\) (Ways to choose 2 men from 4)
Ways\(_3 = \binom{4}{2} \times \binom{3}{1} \times \binom{3}{1} \times \binom{4}{2} = 6 \times 3 \times 3 \times 6 = 324\).
Case 4: \(x = 3\)
If \(x=3\) women are chosen from the man's relatives:
- \(y = 3 - 3 = 0\) men from man's relatives.
- \(z = 3 - 3 = 0\) women from wife's relatives.
- \(w = 3\) men from wife's relatives.
Number of ways = (Ways to choose 3 women from 4) \(\times\) (Ways to choose 0 men from 3) \(\times\) (Ways to choose 0 women from 3) \(\times\) (Ways to choose 3 men from 4)
Ways\(_4 = \binom{4}{3} \times \binom{3}{0} \times \binom{3}{0} \times \binom{4}{3} = 4 \times 1 \times 1 \times 4 = 16\).
Total Number of Invitation Ways
The total number of ways to invite the guests according to the conditions is the sum of the ways calculated for each possible case:
Total Ways = Ways\(_1\) + Ways\(_2\) + Ways\(_3\) + Ways\(_4\)
Total Ways = \(1 + 144 + 324 + 16 = 485\).
| Man's Women (\(x\)) | Man's Men (\(y=3-x\)) | Wife's Women (\(z=3-x\)) | Wife's Men (\(w=x\)) | Calculation (\(\binom{4}{x}\binom{3}{y}\binom{3}{z}\binom{4}{w}\)) | Ways |
|---|---|---|---|---|---|
| \(0\) | \(3\) | \(3\) | \(0\) | \(\binom{4}{0}\binom{3}{3}\binom{3}{3}\binom{4}{0}\) | \(1\) |
| \(1\) | \(2\) | \(2\) | \(1\) | \(\binom{4}{1}\binom{3}{2}\binom{3}{2}\binom{4}{1}\) | \(144\) |
| \(2\) | \(1\) | \(1\) | \(2\) | \(\binom{4}{2}\binom{3}{1}\binom{3}{1}\binom{4}{2}\) | \(324\) |
| \(3\) | \(0\) | \(0\) | \(3\) | \(\binom{4}{3}\binom{3}{0}\binom{3}{0}\binom{4}{3}\) | \(16\) |
| Total Number of Ways | \(485\) | ||||
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Important Questions from Permutation and Combination
The following figure is made with nine squares of length 1 cm each.
A student is standing at point P. The total number of shortest paths to reach point Q from point P, if the student walks along the boundaries of the squares only, is:
Consider the following statements (all the angles are measured in degrees) :
1. If two supplementary angles are $4x + 40$ and $10x$, then $x$ is;
2. The complementary angle of $\frac{4}{5} \times 45$ is :
3. If two complementary angles are in the ratio of $4 : 5$, then the smallest angle is :
Which of the following is correct?
Codes:
1 2 3- There are 40 persons in a palace. If every person shakes hands with every other person, what will be the total number of handshakes?
- How many words can be formed with or without meaning by using all letters of the word DELHI?
A rectangular wall is divided into four squares of equal size where there are two rows each having two squares. The top left square is coloured with green. If, including green, there are three colours available and each square is coloured using any one of these three colours such that no two adjacent squares get painted with the same colour; then how many colour combinations are possible ?
