The factorial of a number \(n\) is defined as the product of all the whole numbers less than \(n\) up to 1.
\(Example 1: Find the factorial of 5 (5 !).
Solution: \(5 !=5 \times 4 \times 3 \times 2 \times 1=120\)
Example 2: Find the number of ways 4 distinct objects can be arranged in a row.
Solution: There are 4 choices for the first distinct object, After selecting the first object, 3 choices are left for the second object, 2 choices for the 3rd object, and so on.
\(\text { Thus, } 4 \text { distinct objects can be arranged in } 4 !=4 \times 3 \times 2 \times 1=24 \text {. }\)Example 3: \(100 ! / 98 !\)
Solution:
\(Example 4: \(\text { Simplify the expression } \frac{n !}{(n-2) !} \text {. }\)
Solution:
\(
\frac{n !}{(n-2) !}=\frac{n(n-1)(n-2) !}{(n-2) !}
\)
Now, we can cancel the common factors:
\(
\frac{n(n-1)(n-2) !}{(n-2) !}=n(n-1)
\)
Example 5: How many ways can you arrange the letters in the word “company” without repeating them?
Solution: For this problem, count the number of letters in the word “company” to find there are six letters. Then, find the factorial of the number six.
\(
6 ! = 6 \times 5 \times 4 \times 3 \times 2 \times 1=720
\)
Now you know that the maximum number of ways, you can arrange the letters in the word “company” with no repeats is 720.
Example 6: What combinations can you make with the colors red, blue, and green?
Solution: Find the factorial for the number three because there are three colors, then list the different combinations.
\(3 ! = 3(3-1)(3-2) = 3 \times 2 \times 1=6\)The six combinations are:
red, blue, green
red, green, blue
green, blue, red
green, red, blue
blue, red, green
blue, green, red
Example 7: How many different positive integers can be formed by rearranging the digits 2, 3, 4, and 5?
Solution: There are 4 choices for the first digit. After selecting the first digit, there are 3 choices left for the second digit, 2 for the 3rd digit, and so on. This gives us \(4 ! = 24\) positive integers.
Example 8: Adam has three balls in a bag – one green, one blue, and one yellow. If he draws the three balls in sequence, what chance is there that he will get the yellow first, the green one second, and the blue one last?
Solution: There are 6 possible sequences in which the balls can be drawn: \(3 !=6\).
There is a chance of 1 over the total number of possibilities to get the yellow-green-blue sequence, so that is \(1 /(3 !)\) or \(1 / 6\) or \(16.7 \%\) chance to get the desired outcome.
Example 9: Four girlfriends want to take a photo together. How many different ways can they stand side by side?
Solution: There are \(4 !=24\) ways.
Example 10: There are seven children to be lined up in a straight line for a photograph
(a) How many different ways are possible?
(b) How many different ways are possible if Sally must be in the middle?
(c) How many different ways are possible if Ahmed is on the far left?
Solution:
(a) As there are 7 children, the number of ways \(= 7 ! = 5040\)
(b) As among 7 one position is fixed (sally must be in the middle), and the other 6 may be chosen to fill the other spots. Thus, we have \(6 ! =720\) ways.
(c) Since the first is fixed (Ahmed’s position is fixed), but the other six are not, we have \(6 ! =720\) ways.
Example 11: \(\text { Evaluate: } \frac{50 !}{48 !}\)
Solution:
\(\frac{50 !}{48 !} = \frac{50 \times 49 \times 48!}{48!}=50 \times 49 = 2450\)Example 12: In how many different ways can the letters of the word ‘FIGHT’ be arranged?
Solution: The given word contains 5 different letters.
Required number of ways 5 != (5 × 4 × 3 × 2 × 1) = 120.
Example 13: In how many different ways can the letters of the word ‘PRESENT’ be arranged?
Solution: The word ‘PRESENT’ contains 7 letters, consisting of 2E and all other 5 are different.
Required number of ways =\(\frac{7 !}{2 !}=2520\).
Explanation: The word PRESENT has 7 letters with E repeated twice. Just pretend that the two E’s are different. Let’s call one of the letters E as \(E_1\) and other as \(E_2\). So the number of ways we can arrange the word PRESET now is \(7!=5040\). However, we over-counted. \(PRE_1SE_2NT\) is the same as \(PRE_2SE_1NT\). To eliminate the extra cases, we need to divide by the ways that the E’s can be arranged, which in this case is just \(2!=2.\)
So Required number of ways =\(\frac{7 !}{2 !}=2520\).
Example 14: How many arrangements can be made out of the letters of the word ‘ENGINEERING’?
Solution: The word ‘ENGINEERING’ contains 11 letters, namely \(3 \mathrm{E}, 3 \mathrm{~N}, 2 \mathrm{G}, 2 \mathrm{I}\) and \(1 \mathrm{R}\).
\(
\begin{aligned}
& \therefore \quad \text { Required number of ways }=\frac{11 !}{(3 !)(3 !)(2 !)(2 !)(1 !)} \\
& \quad=\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 6 \times 2 \times 2 \times 1} \\
& =(11 \times 10 \times 7 \times 6 \times 5 \times 4 \times 3)=277200
\end{aligned}
\)
Example 15: In how many different ways can the letters of the word ‘DESIGN’ be arranged so that the vowels are at the two ends?
Solution: The given word ‘DESIGN’ contains 4 consonants and 2 vowels.
At the two ends, the two vowels can be arranged in 2 ways.
Remaining 4 letters can be arranged in \({4 !}=(4 \times 3 \times 2 \times 1)=24\) ways.
Total number of ways \(=(24 \times 2)=48\).
\(\therefore \quad\) Required number of ways \(=48\).
Example 16: In how many different ways can the letters of the word ‘DAUGHTER’ be arranged so that the vowels always come together?
Solution: The given word contains 8 different letters.
When the vowels AUE are taken together, we may treat them as 1 letter.
Then, the letters to be arranged are DGHTR (AUE)
The vowels can be arranged in \(6 != 720\) ways.
The vowels AUE may be arranged in \(3 != 6\) ways.
Required number of ways \(=(720 \times 6)=4320\) ways.
Example 17: In how many different ways can the letters of the word ‘DIRECTOR’ be arranged so that the vowels are always together?
Solution: In the given word, we treat the vowels IEO as 1 letter.
Thus, we have DRCTR (IEO).
This group has 6 letters in which \(R\) occurs 2 times and other are all different.
Number of ways of arranging these letters \(=\frac{6 !}{2 !}=\frac{(6 \times 5 \times 4 \times 3 \times 2 \times 1)}{2}=360\).
Now, 3 vowels can be arranged among themselves in \(13=6\) ways.
Required number of ways \(=(360 \times 6)=2160\).
Example 18: In how many different ways can the letters of the word ‘DIGEST’ be arranged so that the vowels are never together?
Solution: In the given word DIGEST, we take the vowels IE as one letter.
Then, we can write it as DGST (IE).
This word has 5 letters which can be arranged among themselves in \(5 !=(5 \times 4 \times 3 \times 2 \times 1)=120\) ways
The letters of IE can be arranged in 2 ways.
The number of ways of arranging the letters of given word with vowels together \(=(120 \times 2)=240\) ways.
Number of ways of arranging all the letters of the given word
\(
=6 !=(6 \times 5 \times 4 \times 3 \times 2 \times 1)=720 \text { ways. }
\)
Number of ways of arrangements so that the vowels are never together \(=(720-240)=480\)
Example 19: In how many different ways can the letters of the word ‘DETAIL’ be arranged so that the vowels occupy only the odd positions?
Solution: There are 6 letters in the given word, out of which there are 3 consonants and 3 vowels.
Let us mark these positions as (1) (2) (3) (4) (5) (6).
Now, 3 vowels can be placed at any of 3 places, marked 1, 3,5.
The number of these arrangements \(3 !=6\).
Also, 3 consonants can be placed at the remaining 3 places.
The number of these arrangements \(3 !=6\).
Total number of ways \(=(6 \times 6)=36\)
Example 20: In how many different ways can the letters of the word ABSENTEE be arranged?
Solution: The given word contains 8 letters of which E is taken 3 times.
∴ Required number of ways
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