Outcome is equal to the product of several combinations of multiple variables.

If there are *n* variables of an event then the outcome is equal to the product of combinations of *n* variables.

_{1}×C

_{2}×…×C

_{n}

Remember this,

**Number of Combinations:**

The number of all combinations of *n* things, taken *r* at a time is:

**Examples and Solutions on Combinations of multiple variables**

**Example 1:**

In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

**Solution:**

A team of 3 boys and 3 girls is to be selected from 5 boys and 4 girls.

3 boys can be selected from 5 boys in ^{5}C_{3} ways.

3 girls can be selected from 4 girls in ^{4}C_{3} ways.

Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected

**Example 2:**

Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls each colour.

**Solution:**

There are a total of 6 red balls, 5 white balls, and 4 blue balls.

9 balls have to be selected in such a way that each selection consists of 3 balls of each colour.

Here,

3 balls can be selected from 6 red balls in ^{6}C_{3}, ways.

3 balls can be selected from 5 white balls in ^{5}C_{3}, ways.

3 balls can be selected from 5 blue balls in ^{5}C_{3}; ways.

Thus, by multiplication principle, required number of ways of selecting 9 balls.

**Example 3:**

Find the number of ways of selecting 2 black balls from 5 black balls and 3 red balls from 6 red balls at a time?

**Solution:**

There are 5 black and 6 red balls in the bag.

2 black balls can be selected out of 5 black balls in ^{5}C_{2} ways and 3 red balls can be selected out 0f6 red balls in ^{6}C_{3} ways.

Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls

**Example 4:**

The members of a club are 12 boys and 8 girls. In how many ways can a committee of 3 boys and 2 girls be formed?

**Solution:**

**Example 5:**

In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

**Solution:**

Required number of ways =^{8}C_{5} ×^{10}C_{6} =11760.

**Example 6:** A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?

**Solution:**

Here, order does not matter. Therefore, we need to count combinations. There will be as many committees as there are combinations of 5 different persons taken 3 at a time. Hence, the required number of ways

Now, 1 man can be selected from 2 men in

^{2}C

_{1}ways and 2 women can be selected from 3 women in

^{3}C

_{2}ways. Therefore, the required number of committees

**Example 7:**

In how many ways one chairman, 1 president and 2 vice presidents can be selected from 50 delegates of different countries?

A. 2763600 B. 2764800

C. 2763400 D. 2769200

correct: A, solution:

One chairman can be selected from 50 delegates in ^{50}C_{1} ways.

One president can be selected from remaining 49 in ^{49}C_{1} ways.

Two vice-presidents can be selected from remaining 48 in ^{48}C_{2} ways.

Total number of ways is equal to

^{50}C_{1}×^{49}C_{1}×^{48}C_{2} =2763600.

**Example 8:**

What is the number of both squares and rectangles (of any size) in a chessboard?

A. 204 B. 784 C. 2025 D. 1296

correct: D, solution:

For selecting a rectangle we need to select two horizontal and two vertical lines which can be done in

=^{9}C_{2} ×^{9}C_{2} =1296 ways.

**Example 9:**

Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

**Solution:**

In a deck of 52 cards, there are 4 aces. A combinations of 5 cards have to be made in which there is exactly one ace.

Then, one ace can be selected in ^{4}C_{1} ways and the remaining 4 cards can be selected out of the 48 cards in ^{48}C_{4} ways.

Thus, by multiplication principle, required number of 5 card combinations

**Example 10:**

Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

**Solution:**

From a deck of 52 cards, 5-card combinations have to be made in such a way that in each selection of 5 cards, there is exactly one king.

In a deck of 52 cards, there are 4 kings.

1 king can be selected out of 4 kings in ^{4}C_{1} ways.

4 cards out of the remaining 48 cards can be selected in ^{48}C_{4} ways.

Thus, the required number of S-card combinations is ^{4}C_{1} ×^{48}C_{4}.

**Example 11:**

How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?

**Solution:**

In the word DAUGHTER, there are 3 vowels namely, A, U, and E and 5 consonants, namely, D, G, H, T, and R.

Number of ways of selecting 2 vowels of 3 vowels =^{3}C_{2}=3

Number of ways of selecting 3 consonants out of 5 consonants =^{5}C_{3}=10

Therefore, number of combinations of 2 vowels and 3 consonants =3×10=30

Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! ways.

Hence, required number of different words =30×5!=3600.

**Example 12:**

How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?

**Solution:** In the word INVOLUTE, there are 4 vowels, namely, I,O,E,U and 4 consonants, namely, N, V, L and T.

The number of ways of selecting 3 vowels out of 4 =^{4}C_{3} =4.

The number of ways of selecting 2 consonants out of 4 =^{4}C_{2} =6.

Therefore, the number of combinations of 3 vowels and 2 consonants is 4×6=24.

Now, each of these 24 combinations has 5 letters which can be arranged among themselves in 5! ways. Therefore, the required number of different words is 24×5!=2880.

**Example 13:**

The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?

**Solution:**

Two different vowels and 2 different consonants are to be selected from the English alphabet.

Since there are 5 vowels in the English alphabet, number of ways of selecting 2 different vowels from the alphabet

^{5}C

_{2}=5!/2!3!=10

Since there are 21 consonants in the English alphabet, number of ways of selecting 2 different consonants from the alphabet

^{21}C

_{2}=21!/2!19!=210

Therefore, number of combinations of 2 different vowels and 2 different consonants =10×210=2100

Each of these 2100 combinations has 4 letters, which can be arranged among themselves in 4! ways.

Therefore, required number of words =2100×4!=50400.