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Question: If \(4\) digit numbers not less than \(5000\) are randomly formed from the digits \(0,1,3,5\) and \(...

If 44 digit numbers not less than 50005000 are randomly formed from the digits 0,1,3,50,1,3,5 and 77 , what is the probability of forming a number divisible by 55 when
the digits are repeated?
the repetition of digits is not allowed?

Explanation

Solution

Here two concepts are used-
Probability of an event happening == Number of favorable cases/Number of total cases
Combination is a selection of items from a collection, such that the order of selection does not matter. Formula-
nCr=n!r!(nr)!{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}, where nn is the number of things to choose from and we choose rr from them.
In our question we will see two cases to form a number from a nn number of digits to fill rr number of places,
when digits are repeated=nC1×nC1×nC1×.....×rth = {}^n{C_1} \times {}^n{C_1} \times {}^n{C_1} \times ..... \times rthterm=(nC1)r = {({}^n{C_1})^r}
when digits are not repeated=nC1×n1C1×n2C1×.....n(r1)C1 = {}^n{C_1} \times {}^{n - 1}{C_1} \times {}^{n - 2}{C_1} \times .....{}^{n - (r - 1)}{C_1}

Complete answer:
(i)The number of numbers which are not less than 50005000and formed by 0,1,3,50,1,3,5 and 77when the digits are repeated are-
We have only two digits 55 and 77 that can be at thousand’s place so that the number is not less than50005000.

First we will take 55 as thousand’s place digit-

Now we have 55digits to choose from to fill 33 places so we have to choose 33 times.
Therefore the number of possible ways =53=125 = {5^3} = 125
Then we will take 77 as thousand’s place digit-

And again we have 55digits to choose from to fill 33 places so we have to choose 33 times.
Therefore the number of possible ways =53=125 = {5^3} = 125
Finally the number of numbers which are not less than 50005000and formed by 0,1,3,50,1,3,5 and 77when the digits are repeated are =125+125=250 = 125 + 125 = 250 Statement (1)

(ii)To find the probability of the first case when digits are repeated, this would be the total number of cases.
And we have to find the number of numbers which are not less than 50005000and formed by 0,1,3,50,1,3,5 and 77when the digits are repeated are also divisible by 55.
The divisibility rule for 55 says that the one’s place digit should be 55 or 00.

When a thousand's place digit is 55and the one’s place digit is 55.

Now we have 55digits to choose from to fill 22 places so we have to choose 22 times.
Therefore the number of possible ways =52=25 = {5^2} = 25
When a thousand's place digit is 55and the one’s place digit is 00.

Now we have 55digits to choose from to fill 22 places so we have to choose 22 times.
Therefore the number of possible ways =52=25 = {5^2} = 25
When a thousand's place digit is 77and the one’s place digit is 55.

Now we have 55digits to choose from to fill 22 places so we have to choose 22 times.
Therefore the number of possible ways =52=25 = {5^2} = 25
When a thousand's place digit is 77and the one’s place digit is 00.

Now we have 55digits to choose from to fill 22 places so we have to choose 22 times.
Therefore the number of possible ways =52=25 = {5^2} = 25
Finally the number of numbers which are not less than 50005000and formed by 0,1,3,50,1,3,5 and 77 also divisible by 55when the digits are repeated are =25+25+25+25=100 = 25 + 25 + 25 + 25 = 100
Statement (2)
To find the probability of the first case when digits are repeated, this would be the number of favorable cases.
Probability of forming a number which are not less than 50005000and formed by 0,1,3,50,1,3,5 and 77 also divisible by 55when the digits are repeated
== Number of favorable cases/total number of cases
=100250= \dfrac{{100}}{{250}} [From statement (1) and statement (2)]
=25= \dfrac{2}{5}

(iii)The number of numbers which are not less than 50005000and formed by 0,1,3,50,1,3,5 and 77when the digits are not repeated are-
We have only two digits 55 and 77 that can be at thousand’s place so that the number is not less than50005000.

First we will take 55 as thousand’s place digit-

Now we have 44digits for our first choice, 33 digits for the second and 22 for the third.
Therefore the number of possible ways here=4×3×2=24 = 4 \times 3 \times 2 = 24
Then we will take 77 as thousand’s place digit-

Now we have 44digits for our first choice, 33 digits for the second and 22 for the third.
Therefore the number of possible ways here=4×3×2=24 = 4 \times 3 \times 2 = 24
Finally the number of numbers which are not less than 50005000and formed by 0,1,3,50,1,3,5 and 77when the digits are not repeated are =24+24=48 = 24 + 24 = 48 Statement (3)
To find the probability of a second case when digits are not repeated, this would be the total number of cases.

And we have to find the number of numbers which are not less than 50005000and formed by 0,1,3,50,1,3,5 and 77when the digits are not repeated are also divisible by 55.
The divisibility rule for 55 says that the one’s place digit should be 55 or00.
When a thousand's place digit is 55and the one’s place digit is55.

This case cannot be possible since 55is repeated.
When a thousand's place digit is 55and the one’s place digit is00.

Now we have 33digits for our first choice and 22 digits for the second.
Therefore the number of possible ways here=3×2=6 = 3 \times 2 = 6
When a thousand's place digit is 77and the one’s place digit is 55.

Now we have 33digits for our first choice and 22 digits for the second.
Therefore the number of possible ways here=3×2=6 = 3 \times 2 = 6
When a thousand's place digit is 77and the one’s place digit is 00.

Now we have 33digits for our first choice and 22 digits for the second.
Therefore the number of possible ways here=3×2=6 = 3 \times 2 = 6
Finally the number of numbers which are not less than 50005000and formed by 0,1,3,50,1,3,5 and 77 also divisible by 55when the digits are not repeated are =6+6+6=18 = 6 + 6 + 6 = 18 Statement (4)
To find the probability of a second case when digits are repeated, this would be the number of favorable cases.
Probability of forming a number which are not less than 50005000and formed by 0,1,3,50,1,3,5 and 77 also divisible by 55when the digits are not repeated
== Number of favorable cases/total number of cases
=1848= \dfrac{{18}}{{48}} [from statement (3) and statement (4)]
=38= \dfrac{3}{8}
Hence,
Probability of forming a number which are not less than 50005000and formed by 0,1,3,50,1,3,5 and 77 also divisible by 55when the digits are repeated =25 = \dfrac{2}{5}
Probability of forming a number which are not less than 50005000and formed by 0,1,3,50,1,3,5 and 77 also divisible by 55when the digits are not repeated =38 = \dfrac{3}{8}

Note: The number of ways to arrange nn objects in a row, of which exactly rr objects are identical, isn!r!\dfrac{{n!}}{{r!}}.
In generalize form, if we wish to arrange a total no. of nnobjects, out of which pp are of one type, qq of second type are alike, and rr of a third kind are same, then such a computation is done by n!p!q!r!\dfrac{{n!}}{{p!q!r!}}