+0

0
4194
10

Pat writes all the 7-digit numbers in which all the digits are different and each digit is greater than the one to its right (so the tens digit is greater than the units, the hundreds greater than the tens, and so on). For example, 9,865,320 is one of the numbers that Pat writes down.

(a) How many numbers does Pat write down?

(b) One of Pat's numbers is chosen at random. What is the probability that the tens digit is a 1?

(c) One of Pat's numbers is chosen at random. What is the probability that the middle (thousands) digit is a 5?

Note: of course, you could solve this problem by repeating Pat's experiment and writing down all of the numbers. But don't do that -- figure out the answers without needing to write down all of the numbers!

Apr 3, 2015

#5
+36

Can I make this a shorter answer?    YES I CAN Pat writes all the 7-digit numbers in which all the digits are different and each digit is greater than the one to its right (so the tens digit is greater than the units, the hundreds greater than the tens, and so on). For example, 9,865,320 is one of the numbers that Pat writes down.

(a) How many numbers does Pat write down?

9,8,7,6,5,4,3,2,1,0     3 of the 10 must be removed    10C3 = 120

(b) One of Pat's numbers is chosen at random. What is the probability that the tens digit is a 1?

(9,8,7,6,5,4,3,2) 1,0       3 of  the 8 must be left out     8C3= 56

P(1 in the 10s column) = $${\frac{{\mathtt{56}}}{{\mathtt{120}}}} = {\frac{{\mathtt{7}}}{{\mathtt{15}}}} = {\mathtt{0.466\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

(c) One of Pat's numbers is chosen at random. What is the probability that the middle (thousands) digit is a 5?

(9,8,7,6 one of these four must be left out) 5 ( 4,3,2,1,0 two iof thse 5 must be left out) 4C1*5C2=4*10=40

P(5 is in the middle) =  $${\frac{{\mathtt{40}}}{{\mathtt{120}}}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}} = {\mathtt{0.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

Apr 6, 2015

#1
+12

Results are:

(a) 120

(b) 7/15

(c) 1/3

.

Apr 3, 2015
#2
+12

Thanks Alan,

This looks interesting but finding the time to answer all these counting and probability questions (especially in full)  is becoming very difficult.  I think I speak for more answerers than just myself If you want more explanation try asking for it with a new post.

http://web2.0calc.com/questions/instructions-on-reposting_1

Apr 4, 2015
#3
+12

I agree. That's why I cheated on this one and just wrote a simple program to do the calculations!

.

Apr 4, 2015
#4
+20

Thanks Anon, My instructions for reposting were followed to the T so I am going to give some time to your question.

Alan has already give the answer only version so  that will help me know if my  answers are correct.  Thanks Alan.

---------------

PLEASE NOTE THAT I HAVE DONE IT A MUCH SHORTER WAY IN THE NEXT POST. Pat writes all the 7-digit numbers in which all the digits are different and each digit is greater than the one to its right (so the tens digit is greater than the units, the hundreds greater than the tens, and so on). For example, 9,865,320 is one of the numbers that Pat writes down.

(a) How many numbers does Pat write down?

First I will look at all those beginning in 9

9 8 7 6 5 4 3                 Ends in 3     5 middle digits 0 digits can be left out               5C0=1

9                   2              Ends in 2     6 middle digits any 1 digit must be left out        6C1=6

9                      1           Ends in 1      7 middle digits any 2 digit must be left out       7C2=21

9                          0       Ends in 0       8 middle digits any 3 digit must be left out      8C3=56

Now look at those starting with 8

8 7 6 5 4 3 2                 Ends in 2     5 middle digits 0 digits can be left out                5C0=1

8                   1              Ends in 1     6 middle digits any 1 digit must be left out         6C1=6

8                      0           Ends in 0      7 middle digits any 2 digit must be left out        7C2=21

Now look at those starting with 7

7 6 5 4 3 2 1                Ends in 1     5 middle digits 0 digits can be left out                5C0=1

7                   0              Ends in 0     6 middle digits any 1 digit must be left out         6C1=6

Now look at those starting with 6

6 5 4 3 2 1  0              Ends in 0     5 middle digits 0 digits can be left out                  5C0=1

$${\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}{\mathtt{\,\small\textbf+\,}}{\mathtt{56}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{120}}$$

I agree with Alan, I must be a genious! (b) One of Pat's numbers is chosen at random. What is the probability that the tens digit is a 1?

Mmm

9 -  -  -  - 10    (8,7,6,5,4,3,2 ) 3 out of the 7 will be missing       7C3 = 35

8 -  -  -  - 10    (7,6,5,4,3,2)     2 out of the 6 will be missing       6C2 = 15

7 -  -  -  - 10     (6,5,4,3,2)       1 out of the 5 will be missing       5C1 =  5

6 5 4 3 2 1 0                             0 out of the 4 will be missing       4C0 = 1

$${\mathtt{35}}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{56}}$$

56 of the numbers will have a 1 in the 10s column.

P(1 in the 10s column) = $${\frac{{\mathtt{56}}}{{\mathtt{120}}}} = {\frac{{\mathtt{7}}}{{\mathtt{15}}}} = {\mathtt{0.466\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

WOW that the same as Alan got too - I'm on a roll. (c) One of Pat's numbers is chosen at random. What is the probability that the middle (thousands) digit is a 5?

NOTE:  I have put the 5 in the MIDDLE which is the 10,000 place

_ _ _ 5 _ _ _

9,8,7,6     leave out 1 of these     4C1= 4 ways

4,3,2,1,0   leave out 2 of these    5C2 = 10 ways

4*10 = 40 ways that 5 can be in the middle

P(5 is in the middle) = $${\frac{{\mathtt{40}}}{{\mathtt{120}}}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}} = {\mathtt{0.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

This one does not agree with Alan.    Maybe I am not so clever after all. -----------------------------------------------------------------------

I am thinking that I may have made this question longer than necessary.

I might try and make the working shorter in another post. Apr 6, 2015
#5
+36

Can I make this a shorter answer?    YES I CAN Pat writes all the 7-digit numbers in which all the digits are different and each digit is greater than the one to its right (so the tens digit is greater than the units, the hundreds greater than the tens, and so on). For example, 9,865,320 is one of the numbers that Pat writes down.

(a) How many numbers does Pat write down?

9,8,7,6,5,4,3,2,1,0     3 of the 10 must be removed    10C3 = 120

(b) One of Pat's numbers is chosen at random. What is the probability that the tens digit is a 1?

(9,8,7,6,5,4,3,2) 1,0       3 of  the 8 must be left out     8C3= 56

P(1 in the 10s column) = $${\frac{{\mathtt{56}}}{{\mathtt{120}}}} = {\frac{{\mathtt{7}}}{{\mathtt{15}}}} = {\mathtt{0.466\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

(c) One of Pat's numbers is chosen at random. What is the probability that the middle (thousands) digit is a 5?

(9,8,7,6 one of these four must be left out) 5 ( 4,3,2,1,0 two iof thse 5 must be left out) 4C1*5C2=4*10=40

P(5 is in the middle) =  $${\frac{{\mathtt{40}}}{{\mathtt{120}}}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}} = {\mathtt{0.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

Melody Apr 6, 2015
#6
+11

You are right about the probability of the thousands digit equalling 5 being 1/3 Melody.  I just checked my program and realised I'd put the counter in the wrong place!

.

Apr 6, 2015
#7
+14

DELETED - I'm an idiot.   Only kidding

Thanks Alan :))

Apr 6, 2015
#8
+11

The thousands column is the middle column!

1       2       3                4                  5                          6                                7

units  tens  hundreds  thousands  ten-thousands  hundred-thousands   millions

.

Apr 6, 2015
#9
+11

Wow, Melody....you made that one look easy....very nice........!!!!!!   Apr 6, 2015
#10
+10

Thanks Chris and Alan. :)

It did look a lot harder than it really was :)

Apr 6, 2015