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How many ways are there to put 4 balls into 3 boxes, given that the balls are not distinguished but the boxes are?

 

How many zeroes do we write when we write all the integers from 1 to243  in base 3?

 

In how many ways can 3 identical red balls, 3 identical green balls, and 3 identical blue balls be arranged in a 3 * 3 grid, such that each row and each column of the grid contains 1 ball of each color?

lololololololololll  Jun 28, 2018
 #1
avatar+867 
+4

1. Since the balls are indistinguishable, we just need to keep track of how many balls in each box. 

 

The possible arrangements are:

 

\((4, 0, 0); (3, 1, 0); (2, 2, 0); (2, 1, 1) \Rightarrow \boxed{4}\)

 

 

2. We start by splitting the numbers into three cases:

 

The one digit numbers don't have any zeroes. 

 

The two digit numbers use two zeroes: 10 and 20. 

 

There are \(3^2 = 9\) three-digit numbers starting with 1, and 9 starting with 2. For each leading digit, a zero appears in each digit in \(9\div3 = 3\) of the numbers, so each has a total of \(3 + 3 = 6\) zeroes. Thus, the 3-digit numbers contain \(2\cdot6 = 12 \) zeroes. 

 

3. Here is the 3 by 3 grid:

 

 R   B   G 
 G  R  B
 B  G  R

 

For the first column, we have 3! = 6 arrangements of red, green, and blue. 

For the second column, we have only \(\binom21\) arrangements of red, green, and blue. 

For the last column, we have only 1 arrangements of red, green, and blue. 

 

For a total of 6 * 2 = 12 arrangements. 

 

I hope this helped,

 

Gavin.

GYanggg  Jun 28, 2018
 #2
avatar
+1

Here are all numbers from 1 to 243 in base 3. Do you think you can count all the zeros??

 

{1_3, 2_3, 10_3, 11_3, 12_3, 20_3, 21_3, 22_3, 100_3, 101_3, 102_3, 110_3, 111_3, 112_3, 120_3, 121_3, 122_3, 200_3, 201_3, 202_3, 210_3, 211_3, 212_3, 220_3, 221_3, 222_3, 1000_3, 1001_3, 1002_3, 1010_3, 1011_3, 1012_3, 1020_3, 1021_3, 1022_3, 1100_3, 1101_3, 1102_3, 1110_3, 1111_3, 1112_3, 1120_3, 1121_3, 1122_3, 1200_3, 1201_3, 1202_3, 1210_3, 1211_3, 1212_3, 1220_3, 1221_3, 1222_3, 2000_3, 2001_3, 2002_3, 2010_3, 2011_3, 2012_3, 2020_3, 2021_3, 2022_3, 2100_3, 2101_3, 2102_3, 2110_3, 2111_3, 2112_3, 2120_3, 2121_3, 2122_3, 2200_3, 2201_3, 2202_3, 2210_3, 2211_3, 2212_3, 2220_3, 2221_3, 2222_3, 10000_3, 10001_3, 10002_3, 10010_3, 10011_3, 10012_3, 10020_3, 10021_3, 10022_3, 10100_3, 10101_3, 10102_3, 10110_3, 10111_3, 10112_3, 10120_3, 10121_3, 10122_3, 10200_3, 10201_3, 10202_3, 10210_3, 10211_3, 10212_3, 10220_3, 10221_3, 10222_3, 11000_3, 11001_3, 11002_3, 11010_3, 11011_3, 11012_3, 11020_3, 11021_3, 11022_3, 11100_3, 11101_3, 11102_3, 11110_3, 11111_3, 11112_3, 11120_3, 11121_3, 11122_3, 11200_3, 11201_3, 11202_3, 11210_3, 11211_3, 11212_3, 11220_3, 11221_3, 11222_3, 12000_3, 12001_3, 12002_3, 12010_3, 12011_3, 12012_3, 12020_3, 12021_3, 12022_3, 12100_3, 12101_3, 12102_3, 12110_3, 12111_3, 12112_3, 12120_3, 12121_3, 12122_3, 12200_3, 12201_3, 12202_3, 12210_3, 12211_3, 12212_3, 12220_3, 12221_3, 12222_3, 20000_3, 20001_3, 20002_3, 20010_3, 20011_3, 20012_3, 20020_3, 20021_3, 20022_3, 20100_3, 20101_3, 20102_3, 20110_3, 20111_3, 20112_3, 20120_3, 20121_3, 20122_3, 20200_3, 20201_3, 20202_3, 20210_3, 20211_3, 20212_3, 20220_3, 20221_3, 20222_3, 21000_3, 21001_3, 21002_3, 21010_3, 21011_3, 21012_3, 21020_3, 21021_3, 21022_3, 21100_3, 21101_3, 21102_3, 21110_3, 21111_3, 21112_3, 21120_3, 21121_3, 21122_3, 21200_3, 21201_3, 21202_3, 21210_3, 21211_3, 21212_3, 21220_3, 21221_3, 21222_3, 22000_3, 22001_3, 22002_3, 22010_3, 22011_3, 22012_3, 22020_3, 22021_3, 22022_3, 22100_3, 22101_3, 22102_3, 22110_3, 22111_3, 22112_3, 22120_3, 22121_3, 22122_3, 22200_3, 22201_3, 22202_3, 22210_3, 22211_3, 22212_3, 22220_3, 22221_3, 22222_3, 100000_3}

Guest Jun 28, 2018
 #3
avatar+19624 
+2

2.

How many zeroes do we write when we write all the integers from 1 to243  in base 3?

 

\(\text{Let base $b = 3$} \\ \text{Let digit numbers $= n$ }\)

 

1. The one digit numbers \((n=1)\) don't have any zeroes. 

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 0\times \dfrac{2\cdot 3^0}{3} = 0 \text{ zeroes}\)

 

2. The two digit numbers \((n=2)\) have:

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 1\times \dfrac{2\cdot 3}{3} = 2 \text{ zeroes}\)

 

3. The three digit numbers \((n=3) \) have:

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 2\times \dfrac{2\cdot 3^2}{3} = 12 \text{ zeroes}\)

 

4. The four digit numbers \((n=4)\) have:

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 3\times \dfrac{2\cdot 3^3}{3} = 54 \text{ zeroes}\)

 

5. The five digit numbers \((n=5)\) have:

\((n-1)\times \dfrac{(b-1)\cdot b^{(n-1)} }{b} = 4\times \dfrac{2\cdot 3^4}{3} = 216 \text{ zeroes}\)

 

6. The six digit numbers \((n=6) \) have:

\(243 = 100000_3 = 5 \text{ zeroes}\)

 

\(\text{Ths sum $= 0 + 2 + 12 + 54 + 216 + 5 = 289$ zeroes } \)

 

We write 289 zeroes, when we write all the integers from 1 to 243  in base 3

 

laugh

heureka  Jun 28, 2018
edited by heureka  Jun 28, 2018
 #4
avatar+87301 
+3

How many ways are there to put 4 balls into 3 boxes, given that the balls are not distinguished but the boxes are?

 

Note that the boxes are distinguishable....so....(4, 0, 0)  is different from (0, 4 , 0)

 

The  total number of ways  =  

 

C ( n + k - 1, k)     where  n = the number of boxes   and  k  = the number of balls

 

So we have

 

C ( 3 +  4  - 1 , 4)  = C (6 , 4)  = 15  ways

 

We can confirm this by noting that

 

(4, 0, 0)   has  C(3,1)   =  3 arrangements

(3,1,0)  has 3!  = 6 arrangements

(2,2, 0)  has C (3,1) = 3 arrangements

(2, 2 , 1)  has C(3,1)  = 3 arrangements

 

So

 

3 + 6 + 3  + 3   =   15  arrangements (ways)

 

 

cool cool cool

CPhill  Jun 28, 2018

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