Have a question?
Message sent Close
Notifications
Clear all

Counting

11 Posts
4 Users
1 Reactions
3,774 Views
0
Topic starter

How many 5 digit numbers start with a prime number and end with a prime number? Further, no other digit in the number is a prime number. Also, no two digits of the 5 digit number are same .

aniket prajapati 04/10/2018 3:45 pm
15386479993291357462123

is it right sir ?

5 questions & discussions are there under this sub-topic
1
1D3D3FE9 06A5 4EFE 8F88 EB40E11F5159

solution please

aniket prajapati 15/10/2018 8:35 pm

The center of the coin is at least one coin radius from any grid line.
You can then shade the area of a grid square where the coin center cannot fall -- this shaded area will look the same in every square. If the coin radius is r and the grid square sides have length a, there's a small square of side length a−2r in each square where the center can fall without the coin extending beyond the grid square.

So the probablity of staying within the square becomes    

(a-2r)2/a

putting the given values probability=64/100=16/25

0
pnc
TG Team Topic starter 03/10/2018 1:09 am

Divisible by 12 means divisible by 3 and 4 both. 

to be divisible by 4 last two digits has to be divisble by 4. only combination : 76 

_ _ _ _ _ _ _ 76 

to be divisble by 3 sum of the digits has to be divisible by 3 . 

sum of last two digits = 7 + 6 i.e 13 ( 3k + 1) 

so sum of remaining 7 digits should be of the form 3k + 2 . 
_ _ _ _ _ _ _ 
All sevens => sum 49 ( 3k + 1) hence not divisble by 3 . 

6 sevens , 1 sixes => sum 48 ( 3k ) not divisble by 3. 

5 sevens , 2 sixes => sum 47 ( 3k + 2) divisble by 3. 

7777766 => 7!/(5!2!) numbers 

4 sevens , 3 sixes => not divisble by 3

3 sevens , 4 sixes => not divisble by 3 

2 sevens , 5 sixes => divisble by 3. 
6666677 => 7!/(5!2!) numbers 

1 seven , 6 sixes => not divisble by 3 
all sixes => not divisible by 3 

Hence , 2 x 7!/(5!2!) = 42  , 9 digit numbers 

0
prob
aniket prajapati 05/10/2018 8:37 pm

Jaune Y = (4 C1)*(5 C 2) = 4*10 = 40
Jaune X = (4 C 2)*(5 C 1) + (4 C 3) = 6*5 + 4 = 34
Total combinations = 9C 3 = 84

Probability of Jaune = (40 + 34)/84 = 37/42

aniket prajapati 05/10/2018 10:24 pm

Alternate approach,

If the new colour doesn't contain Blue paint, then it isn't of type Jaune. So, it is much easier to calculate the complementary probability.
P(no Blue paint chosen) = (5/9)*(4/8)*(3/7) = 5/42.
Therefore, the required probability is 1 - 5/42 = 37/42

0
probablity

Sir, please help.

aniket prajapati 05/10/2018 8:33 pm

if we assume that each couples as a one person total 3 person in a row and the probablity of three person, where couples seat together, on five chairs is 3/5.

But the question is asking neither of couples seat together so the opposite way 1-3/5=2/5

0
6DDCCBEA B0BA 41BF A984 9ECC972D3135

any other approach to this TG solved example? Unable to comprehend the given solution.