# How to calculate combinations of multiple item types

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.

outcome=C1×C2×…×Cn

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 5C3 ways.
3 girls can be selected from 4 girls in 4C3 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 6C3, ways.
3 balls can be selected from 5 white balls in 5C3, ways.
3 balls can be selected from 5 blue balls in 5C3; 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 5C2 ways and 3 red balls can be selected out 0f6 red balls in 6C3 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 =8C5 ×10C6 =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 2C1 ways and 2 women can be selected from 3 women in 3C2 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 50C1 ways.
One president can be selected from remaining 49 in 49C1 ways.
Two vice-presidents can be selected from remaining 48 in 48C2 ways.
Total number of ways is equal to
50C1×49C1×48C2 =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
=9C2 ×9C2 =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 4C1 ways and the remaining 4 cards can be selected out of the 48 cards in 48C4 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 4C1 ways.
4 cards out of the remaining 48 cards can be selected in 48C4 ways.
Thus, the required number of S-card combinations is 4C1 ×48C4.

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 =3C2=3
Number of ways of selecting 3 consonants out of 5 consonants =5C3=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 =4C3 =4.
The number of ways of selecting 2 consonants out of 4 =4C2 =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

5C2=5!/2!3!=10

Since there are 21 consonants in the English alphabet, number of ways of selecting 2 different consonants from the alphabet
21C2=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.

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