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avatar+178 

1) How many distinct odd 4-digit numbers can be written with the digits 1, 2, 3 and 4 if no digit may be used more than once?

 

2 )License plate numbers in Aopslandia used to consist of 5 distinct digits. Due to an increase in population, Aopslandia started to allow license plates to have any 5-digit string, even if some digits are the same. How many more license plates are available because of this change?

 

3) How many distinct arrangements can be made from the letters in the word "REARRANGE''?

 

4) Each of the digits 1,2,3,4,5,6 is used exactly once in forming the 3-digit integers X and Y. How many possible values of X+Y are there if |X-Y|=111?

 

5) The product of all digits of positive integer M is 105. How many such Ms are there with distinct digits?

 

6) Two boys and three girls are going to sit around a table with 5 different chairs. If the two boys want to sit together, in how many possible ways can they be seated?

 

7) What fraction of all the 10-digit numbers with distinct digits have the property that the sum of every pair of neighboring digits is odd?

 

Thanks!

 Apr 24, 2019
 #1
avatar+5797 
+2

1) The last digit must be 1 or 3.  The rest of the digits can then be arranged in 3! ways.

 

Thus there are 2x3! = 12 odd 4 digit numbers using digits 1-4

 Apr 24, 2019
 #5
avatar+178 
+2

Thanks!

Rudram592  Apr 24, 2019
 #2
avatar+84 
0

2) First we calculate how many possibilities before: 

\(9\cdot8\cdot7\cdot6\cdot5=15120\)

 

Now we calculate the possibilites now:

\(9\cdot9\cdot9\cdot9\cdot9=9^5=59049\)

 

We can get the difference:

 

\(59049-15120=\boxed{43929}\)

.
 Apr 24, 2019
 #3
avatar+84 
+1

3) In the word REARRANGE we have 9 letters.

 

For these letters to be distinct, we must think of each letter like a gumball in a gumball machine. Every time I use one,  there of less to be used.

 

We also need casework for each length of word:

CASE 1: 1 letter words

 

 one letter words.

CASE 2: 2 letter words

 

\(9\cdot8=72\) two letter words

CASE 3: 3 letter words.

 

\(9\cdot8\cdot7=504\)

CASE 4: 4 letter words

 

\(9\cdot8\cdot7\cdot6=3024\)

CASE 5: 5 letter words

 

\(9\cdot8\cdot7\cdot6\cdot5=15120\)

CASE 6: 6 letter words

 

\(9\cdot8\cdot7\cdot6\cdot5\cdot4=60480\)

CASE 7: 7 letter words (almost done!)

 

\(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3=181440\)

CASE 8: 8 letter words

 

\(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2=362880\)

CASE 9: (DONE!)

 

\(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=362880\)

Now we add all the cases together for the total number of arrangments:

 

\(9+72+504+3024+15120+60480+181440+2(362880)=\boxed{986409}\)

 

 

 

SIDENOTE:

 

You can subsitute the decending numbers with a factorial

 

\(5\cdot4\cdot3\cdot2\cdot1=5!\)

 

\(5\cdot4\cdot3=\displaystyle\frac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1}=\displaystyle\frac{5!}{2!}\)

 

Just some rules for future reference, hope you enjoy and use them.

 

As always if you have any questions just ask.

 

Goodnight! 🌙

 Apr 24, 2019
edited by Bxtterman  Apr 24, 2019
 #4
avatar+178 
+1

Thanks Bxtterman! You really helped me understand the question and the rules for factorials!

Rudram592  Apr 24, 2019
 #7
avatar+5797 
0

this isn't quite right.

 

You've neglected to take into account the fact that some of the letters are repeated.

Rom  Apr 24, 2019
 #8
avatar+84 
+1

How are they repeated?

Bxtterman  Apr 24, 2019
 #9
avatar+84 
+1

Wait, is this AoPS week 24?!!!!!!!! I'm taking the same class as you I think!

Bxtterman  Apr 24, 2019
 #10
avatar+5797 
0

there are 3 R's, 2 E's, 2 A's

 

some of the permutations will be identical because of this

Rom  Apr 24, 2019
 #6
avatar+5797 
+1

7) 

\(\text{In order for every pair of neighboring digits to sum to an odd number }\\ \text{the parity of the digits must strictly alternate}\\ \text{There are 4 ways to choose an even first digit}\\ \text{Then there are }5^9 \text{ ways to complete that 10 digit number alternating parity}\\ \text{There are 5 ways to choose an odd first digit}\\ \text{and then again }5^9 \text{ ways to complete the number}\\ \text{This gets us a total of }9\cdot 5^9 \text{ total numbers that strictly alternate digit parity}\\ 9 \cdot 5^9\)

 

\(\text{There are a total of }9\cdot 10^9 \text{ 10 digit numbers}\\ p = \dfrac{9\cdot 5^9}{9\cdot 10^9} = \left(\dfrac 1 2\right)^9 = \dfrac{1}{512}\)

.
 Apr 24, 2019
 #11
avatar+103148 
+1

5) The product of all digits of positive integer M is 105. How many such Ms are there with distinct digits?

 

Here's my best attempt.....

 

105 factors as  3 * 5 * 7

 

So...we have the following possibilities

 

357    

375

537

573

735

753

 

So...6 M's

 

 

cool cool cool

 Apr 24, 2019

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