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How many even five-digit numbers contain each of the digits 1 through 5?
The restriction requires that the last digit be either a 2 or a 4. This allows two choices for the units digit. After selecting either a 2 or a 4 , there are 4 remaining digits for the tens place, 3 for the hundreds, 2 for the thousands, and 1 for the ten thousands place:
\(
2(4 !)=2 \cdot 4 \cdot 3 \cdot 2 \cdot 1=48
\)
How many of the arrangements of the letters in the word COUNTING contain a double-N?
The restriction requires that the N’s be grouped together, so we do just that: treat them as a single letter. Now we just count the number of ways to arrange the letters COU(NN)TIG. There are 7 letters, so we get \(P(7,7)=7 !=5,040\).
In how many of the arrangements of the letters in the word EXAMPLE are the letters A and M adjacent to each other?
However, the letters \(A\) and \(M\) may appear as AM or MA. Additionally, we have “letters” with two E’s. Group the A and the M. This gives us six letters with two E’s. There are 6 ! ways to arrange these, but we must divide by 2 to account for the two E’s. Then we multiply by 2 because the (AM) can be ordered AM or MA. This gives us right back to \(\)6 !=720\(\).
Determine the number of arrangements of these six elements: a, a, a, b, b, c.
\(
n=\frac{6 !}{3 ! \cdot 2 !}=60
\)
How many characters can we create from two commas and four dots?
\(
n=\frac{6 !}{2 ! \cdot 4 !}=15
\)
How many three-lettered words can be formed from letters A B C D E G H if repeats are not allowed?
\(
n=7 \cdot 6 \cdot 5=210
\)
How many different four-letter words can we create from the letters of the word JAMA?
\(
n=\frac{4 !}{2 !}=12
\)
Sam has 4 identical yellow dice and 3 identical blue dice. How many different colored snakes can make them?
\(
n=\frac{(3+4) !}{3 ! \cdot 4 !}=35
\)
How many words can be created from the phrase MATEMATIKA by changing the letters’ order, regardless of whether the words are meaningful?
\(
n=\frac{10 !}{3 ! \cdot 2 ! \cdot 2 ! \cdot 1 ! \cdot 1 ! \cdot 1 !}=151200
\)
Calculate the number of ways of placing four black balls, four turquoise balls, and five gold balls in a row.
\(
n=\frac{(a+b+c) !}{a ! \cdot b ! \cdot c !}=\frac{(4+4+5) !}{4 ! \cdot 4 ! \cdot 5 !}=90090
\)
We want to plant five fruit trees in the garden, of which three are apple trees and two pears. How many different ways can we organize them?
\(
n=\frac{5 !}{3 ! \cdot 2 !}=10
\)
How many three-digit numbers do we make from the numbers 4,5,6,7?
\(
n=4^3=64
\)
Three boys and four girls. How many ways can they be placed side by side according to gender?
\(
n=\frac{(3+4) !}{3 ! \cdot 4 !}=35
\)
The father wants to plant two seedbeds of carrot and two seedbeds of onion. Use a tree chart to find how many different options for placing the seedbeds he has.
\(
n=\frac{4 !}{2 ! \cdot 2 !}=6
\)
How many three-digit numbers are from the numbers 0 2 4 6 8 (with/without repetition)?
\(
\begin{aligned}
& x=4 \cdot 5 \cdot 5=100 \\
& y=4 \cdot 4 \cdot 3=48
\end{aligned}
\)
How many ways can we assemble five wagons when sand is in three wagons and cement in two?
\(
n=\frac{5 !}{3 ! \cdot 2 !}=10
\)
The arranger is to display three identical beige, two identical green, and one black coat in the shop window. How many ways can it do that?
\(
n=\frac{(3+2+1) !}{3 ! \cdot 2 ! \cdot 1 !}=60
\)
How many ways can we divide nine workers into three workplaces if they need four workers in the first workplace, 3 in the second workplace, and 2 in the third?
\(
n=\frac{9 !}{4 ! \cdot 3 ! \cdot 2 !}=1260
\)
Find the number of possible different arrangements of the letters of the word OPTICAL such that the vowels would always be together.
\(
\begin{aligned}
& \text { OIA } \\
& \text { PTCL } \\
& n_1=(4+1) !=120 \\
& n_2=3 !=6 \\
& n=n_1 \cdot n_2=120 \cdot 6=720
\end{aligned}
\)
For the pick 3 lottery, six balls numbered 1 through 6 are placed in a hopper and randomly selected one at a time without replacement to create a three-digit number. How many different three-digit numbers can be created?
The first digit can be any of the 6 numbered balls, after which there are 5 left and then 4 remaining for the last draw., for a total of 6 . 5 . 4 = 120 different 3-digit numbers.
How many ways can five different books be ordered on a shelf from left to right?
5 books can be placed 1st, then 4, 3, 2, and 1. 5!=120.
The eight members of student council are asked to select a leadership team: president, vice-president, and secretary. How many different leadership teams are possible?
\(
{ }^8 \mathbf{P}_{\mathbf{3}} = 8. 7. 6 = 336
\)
You are ranking your elective class choices for the upcoming school year. You must rank your top 5 selections in order of preference. How many ways can you rank 5 of the 7 electives that interest you?
There are 7 choices for your top pick, then 6, 5, 4, and 3 choices: 7. 6. 5. 4. 3=2520 rankings.
Find the number of arrangements of the letters in the word LADDER.
LADDER has 6 letters with two D’s: 6!/2!=360 arrangements.
Find the number of arrangements of the letters in the word ICICLE.
ICICLE has 6 letters with two I’s and two C’s.
Number of arrangements = 6!/(2!. 2!) =180
How many arrangements of the letters in the word FOOT include a double-O?
Consider the two O’s as one letter, so we have F[OO]T has 3 objects and so we have 3!=6 ways to arrange them and each arrangement will include a double-O.
How many arrangements of the letters in the word START begin with a T?
If we place T at the beginning of the word, we are left with 4, 3, 2, and 1 letter to place 2nd through 5th position. Thus there are 4. 3. 2. 1=24 letter arrangements.
How many arrangements of the letters in the word BEGIN start with a vowel?
There are two choices for the first letter (either I or E), after which there are 4 letters available which give 4!=24 arrangements. This gives a total of 2(4!) = 48 possible arrangements.
How many arrangements of the letters in the word BEGINNING have an N at the beginning?
As N is fixed at the beginning we have rest 8 letters with two N’s, I’s, and G’s. So we have 8!/(2!. 2! . 2!) = 5040 arranements.
How many of the distinct arrangements of the digits 1 through 6 have the 1 to the left of the 2?
There are 6!=720 arrangements of the 6 digits. In these 720 arranm=gements, each can be paired with its palindrome(the same number written in reverse order, like 235, 146, and 641, 532). Only one of the two will have the 1 to the left of the 2, so we need to divide 720 by 2 to get 360 arrangements.
Seven students line up on stage. If Molly insists on standing next to Katie, how many different ways are there to arrange the students on stage from left to right?
Consider Molly and Katie as one person and arrange the seven students as if they were 6. There are 6!=720 ways to do this. Additionally, Katie and Molly can stand in two orders (either KM or MK), doubling the number of arrangements to 1440.
How many arrangements of the letters in the word ORDERED include the word RED?
Let’s consider RED as a single letter. Now we have a total [RED]ORDE, 5 letters having 5!=120 ways. However, we over-count three situations where the word RED appears twice: REDREDO, REDORED, and OREDRED (for example, [RED]REDO is the same as [RED]REDO). Subtract these three over-counts to get 117 ways.
How many arrangements of the letters in the word ORDERED begin and end with the same letter?
There are two D’s, two E’s, and two R’s. For the word beginning with D and ending with D, we have to arrange 5 more letters, two of which are E’s and two of which are R’s: 5!/(2! . 2!) = 30 words which begin and end with D. There are an equal number which begins and end with E and with R for a total of 30+30+30 = 90 arrangements.
There are nine parking spots in front of the building for six teachers and the principal. If the principal always gets one of the three shady spots, how many ways can all seven cars be parked in the lot?
First, place the principal in one of the three shady spots. This leaves 8 spaces for 6 teachers. There are \(
{ }^8 \mathbf{P}_{\mathbf{6}}=8 . 7. 6. 5. 4. 3= 20160\) ways for the remaining teachers to park. Multiply this by the three places the principal can park and we get 60480 ways.
Find the number of arrangements of the letters in the word COOKBOOK.
COOKBOOK has 8 letters with four O’s and two K’s. So the number of arrangements = 8!/(4! . 2!) = 840.
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