# Why these two permutations aren't same?

Why does the pair of permutations are not identical in below cases?

Case-1:

a) Number of ways in which 5 girls and 5 boys can be arranged in a row if no two boys are together? $$(6 \cdot 5!)$$

b) Number of ways in which 5 girls and 5 boys can be arranged in a row such that the girls and boys are alternate? $$(2 \cdot 5! \cdot 5!)$$

Case-2:

a) - number of ways in which five different rings can be worn in four fingers with at least one ring in each finger? ($$480$$)

b)- number of ways in which five different rings can be put on four fingers with at least one ring on each finger. ($$960$$)

The numbers in the respective brackets are the textbook answers. Please help to distinguish the pairs . From many days, I am confused on these.

• The answer to part (a) of case 1 should be $6!5!$, not $6 \cdot 5!$. Please check the wording for case 2. If the wording is correct in both cases, then the answer to part (b) is wrong. Mar 2, 2022 at 11:18
• The answer to part (a) is 6*5! And all wordings are correct Mar 2, 2022 at 12:40
• The answer to part (a) can be expressed as $6 \cdot 5! \cdot 5!$. If the answer does not include two factors of $5!$, it is wrong. Mar 2, 2022 at 12:43
• In the textbook , solution goes as following :. In the question, there is no condition for arranging the girls . Now , girls can be arranged in 5! Ways . When girls are arranged , six gaps are generated in which five boys can be arranged . Hence , total arrangements= 5! * 6P5 = 5! * 6 Mar 2, 2022 at 12:48
• I can understand the solution. But my problem is that how to distinguish it's paired situation. Mar 2, 2022 at 12:50

Case 1: It is possible that no two boys are together even when boys and girls are not in alternating positions since girls may be adjacent when no two boys are together but not when boys and girls alternate. Let $$b$$ represent the position of a boy; let $$g$$ represent the position of a girl. There are $$6$$ ways to arrange five $$b$$s and five $$g$$s so that no two $$b$$ are adjacent.

$$bgbgbgbgbg$$

$$bgbgbgbggb$$

$$bgbgbggbgb$$

$$bgbggbgbgb$$

$$bggbgbgbgb$$

$$gbgbgbgbgb$$

Notice that boys and girls alternate only in the first and last of these six arrangements.

There are $$5!$$ ways to arrange the boys in the five positions occupied by $$b$$s and $$5!$$ ways to arrange the girls in the five positions occupied by $$g$$s.

Therefore, there are $$6 \cdot 5!5! = 6!5!$$ arrangements of five boys and five girls in which no two boys are together and $$2 \cdot 5!5!$$ arrangements of five boys and five girls in which boys and girls and alternate.

Case 2: The two questions are equivalent. The first of the two answers is correct.

To make the problem concrete, suppose the fingers on which the rings are to be placed are the four fingers of the left hand (not including the thumb). We can do this since the four fingers, while not specified, are given.

Since five rings are placed on four fingers, exactly one of the four fingers must receive two rings, while each of the other fingers receives one ring apiece. There are four ways to select the finger which will receive two rings.

For each of these four distributions, there are $$5!$$ ways to arrange the rings from the bottom left on the pinky finger to the top right on the index finger. Hence, the number of ways five distinct rings can be placed (or worn) on four fingers if at least one ring is placed on each finger is $$4 \cdot 5! = 480$$

In the comments, you said the book's solution is:

Since at least one ring is to be put in each finger. So, we select four rings out of $$5$$ and then they are arranged in fingers. This can be done in $$5C4 \cdot 4!$$ ways. Now, one ring is left out which can be put on finger in two ways, either above the ring already placed on it or below it. So, the required number of ways is $$= 5C4 \cdot 4! \cdot 4 \cdot 2 = 960$$ ways.

This answer is incorrect. Suppose the two rings which are placed on the same finger are a gold band and a diamond ring. There are only two ways to arrange these rings on one finger. Either the gold band is above or below the diamond ring.

The authors count the arrangement with a gold band on the bottom and a diamond ring on the top in two ways:

1. The gold band is placed on the finger first, then the diamond ring is the additional ring which is placed on the finger above the gold band.
2. The diamond ring is placed on the finger first, then the gold band is the additional ring which is placed on the finger below the diamond ring.

The authors count the arrangement with a diamond ring on the bottom and a gold band on the top in two ways:

1. The diamond ring is placed on the finger first, then the gold band is the additional ring which is placed on the finger above the diamond ring.
2. The gold band is placed on the finger first, then the gold band is the additional ring which is placed on the finger below the gold band.

The authors of your book overlooked the fact that the first ring placed on a finger must be placed at the bottom of the finger. Consequently, they count every arrangement twice. To correct that error, you must divide their answer by $$2$$ to obtain the correct answer of $$480$$ possible arrangements.

• Thanks , I understood the difference between the situations in case 1 . However , in case 2 , It's not allowed to put/worn 2 fingers at once in same finger . Please read question again Mar 2, 2022 at 13:07
• Moreover , there is no restriction on thumb Mar 2, 2022 at 13:08
• In order to place five rings on four fingers, you must place two rings on one finger and one ring apiece on each of the other three fingers. Choosing to place the rings on the four fingers of the left hand is just an example to make the problem concrete. It does not matter which four fingers are used. If you had to choose on which four fingers the rings should be placed, you would have to multiply the answer by $\binom{10}{4} = 210$. Mar 2, 2022 at 13:16
• No , actually we have to put/worn 5 rings on any 4 fingers of a single hand (not two hands) . I Mar 2, 2022 at 13:20
• Again, we are given the four fingers. I used the four fingers of the left hand as an example in order to explain how I was distributing the rings to the finger from bottom left to top right. I have revised my answer to make the double counting in the author's solution more explicit. Mar 3, 2022 at 10:14