Understanding "Out of p + q + r + t things p, q, r and t are alike"

In the study of permutations and combinations, a common type of problem involves counting the number of ways to arrange or select items when there are groups of identical items among them. The phrase "Out of p + q + r + t things p, q, r and t are alike" refers to a scenario where we have a total of p + q + r + t items, but within this total, there are groups of items that are identical to each other. Specifically, there are p items that are identical to each other, q items that are identical, r items that are identical, and t items that are identical.

Permutations of Identical Items

When arranging items, the order matters. If all items were distinct, the number of permutations would be the factorial of the total number of items. However, when there are identical items, we must adjust the formula to account for the indistinguishability of these items.

The formula for the number of permutations of n items where there are groups of identical items is:

$$ P = \frac{n!}{p! \cdot q! \cdot r! \cdot t!} $$

where:

• ( n = p + q + r + t ) is the total number of items,
• ( p! ) is the factorial of the number of identical items in the first group,
• ( q! ) is the factorial of the number of identical items in the second group,
• ( r! ) is the factorial of the number of identical items in the third group,
• ( t! ) is the factorial of the number of identical items in the fourth group.

Table of Differences and Important Points

Aspect	Distinct Items	Identical Items
Total Items	( n )	( p + q + r + t )
Permutation Formula	( n! )	( \frac{(p + q + r + t)!}{p! \cdot q! \cdot r! \cdot t!} )
Consideration of Order	Order matters	Order of identical items does not produce a unique arrangement
Example	Arranging 5 distinct books on a shelf	Arranging 5 books where 2 are identical copies of one title, and 3 are identical copies of another

Examples

Example 1: Simple Arrangement

Suppose we have 5 balls, where 2 are red (p = 2) and 3 are blue (q = 3). How many different ways can we arrange these balls in a line?

Using the formula:

$$ P = \frac{(p + q)!}{p! \cdot q!} = \frac{(2 + 3)!}{2! \cdot 3!} = \frac{5!}{2! \cdot 3!} = \frac{120}{2 \cdot 6} = \frac{120}{12} = 10 $$

There are 10 different ways to arrange these balls.

Example 2: Complex Arrangement

Imagine we have a collection of 20 fruits: 5 apples (p = 5), 7 bananas (q = 7), 4 oranges (r = 4), and 4 pears (t = 4). In how many different ways can we arrange these fruits in a row?

Using the formula:

$$ P = \frac{(p + q + r + t)!}{p! \cdot q! \cdot r! \cdot t!} = \frac{(5 + 7 + 4 + 4)!}{5! \cdot 7! \cdot 4! \cdot 4!} $$

We calculate the factorials and simplify:

$$ P = \frac{20!}{5! \cdot 7! \cdot 4! \cdot 4!} $$

This calculation would typically be done with a calculator or computer software due to the large numbers involved.

Conclusion

When dealing with permutations of items where some are identical, it is crucial to adjust the formula to account for the indistinguishability of these items. The key takeaway is that the order of identical items does not contribute to unique arrangements, which is why we divide by the factorials of the number of identical items. Understanding this concept is essential for solving problems in permutations and combinations involving identical items.