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A 5 digit number is formed by using the digits 0, 1, 2, 3, 4 and 5 without repetition. The probability that the number is divisible by 6 is?
$A.$ 8%
$B.$ 17%
$C.$ 18%
$D.$ 36%
Answer
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Verified
Verified
Hint: In this question first find out the total number of 5 digit numbers later on find the numbers which are divisible by 6 later on use the property of probability to reach the solution of the question.
Complete step-by-step answer:
Given digits 0, 1, 2, 3, 4 and 5.
Now we have to make a 5 digit number.
So, the number of ways to select 5 digit out of 6 is ${}^6{C_5}$ which is equal to 6 $\left[ {\because {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}} \right]$
Now we have to arrange these numbers. So the possible arrangement is (5!).
So, the possible 5 digit numbers are $\left( {6 \times 5!} \right)$.
Now if the number is starting from 0 it is not a five digit number it is a four digit number so we have to subtract all four digit numbers from the possible five digit numbers.
So all the possible combination of five digits are (0, 1, 2, 3 and 4), (0, 1, 2, 3 and 5), (0, 1, 2, 4 and 5), (0, 1, 3, 4 and 5), (0, 2, 3, 4 and 5), (1, 2, 3, 4 and 5).
So out of these possible cases there are 5 possible cases in which numbers can start from 0.
So number of four digits numbers are $\left( {5 \times 4!} \right)$. (when the number is
starting from 0 the possible arrangements of the remaining digits is (4!))
So, the total 5 digit numbers are $ = \left[ {\left( {6 \times 5!} \right) - \left( {5 \times 4!}
\right)} \right] = \left[ {6 \times 120 - 5 \times 24} \right] = 720 - 120 = 600$
Now we have to find out the 5 digit numbers which are divisible by 6.
So we all know that if a number is divisible by 6 it is also divisible by 2 and 3.
So if a number is divisible by 3 the sum of all the digits should also be divisible by 3.
So the sum of all the digits is (0 + 1 + 2 + 3 + 4 + 5 = 15) so fifteen is divisible by 3.
Now we have to consider only 5 digits so there are two possible cases.
Case – 1 (0, 1, 2, 4, 5)
Case – 2 (1, 2, 3, 4, 5)
(Because if we subtract 0 or 3 the sum of digits is again divisible by 3 and when we subtract any other digit the sum of the digits is not divisible by 3)
Now consider case 1.
Case – 1 (0, 1, 2, 4, 5)
Now we have to find the numbers which are also divisible by 2.
As we know the number is divisible by 2 if the last digit of the number is even or 0.
So if the last digit is zero (0) so the number of possible ways which is divisible by 6 is = (4!).
$ \Rightarrow 4! = 4 \times 3 \times 2 \times 1 = 24$
Now if the last digit is not 0, so for this case there are two possible ways (2, 4)
And we all know numbers cannot start from 0. So, the first digit is chose from the remaining numbers
So, from (2, 4) one digit is placed as last digit and 0 cannot be the first digit so the remaining digits are (5 – 2 = 3)
So the number of ways to choose first digit is ${}^3{C_1} = 3$
So, first digit and last digit are chosen, now we have to make the arrangement for the middle three digits.
So, the possible number of arrangements is = (3!).
So, the numbers which are divisible by 6 and not starting from 0 are = $\left( {{}^3{C_1}
\times 3! \times 2} \right) = 3 \times 6 \times 2 = 36$
So, the total number of 5 digit number which is divisible by 6 for case 1 is (24 + 36 = 60)
Now consider case 2
Case -2 (1, 2, 3, 4, 5)
In this case there is no zero, so the last digit is filled by again in two possible ways (2, 4).
And the remaining four numbers can permute in (4!) ways.
So, the numbers which is divisible by 6 = $\left( {2 \times 4!} \right) = 48$
So, the total numbers including case 1 and case2 which are divisible by 6 = (60 +48 = 108).
So, the required probability (p) that the number is divisible by 6 is
\[ \Rightarrow p = \dfrac{{{\text{Favorable number of outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{108}}{{600}} = \dfrac{{18}}{{100}} = 0.18\]
Now we have to convert this probability into percentage so we have to multiply by 100.
$ \Rightarrow p = 100\left( {0.18} \right) = 18$%.
So, this is the required answer.
Hence, option (c) is correct.
Note: Whenever we face such types of questions the key concept we have to remember is that the probability is the ratio of favorable number of outcomes to the total number of outcomes so, first find out the total number of outcomes as above then find out the favorable number of outcomes using the concept of divisibility rule of 6 which is stated above then divide these according to probability theorem and convert it into percentage by multiplying 100, we will get the required probability in percentage.
Complete step-by-step answer:
Given digits 0, 1, 2, 3, 4 and 5.
Now we have to make a 5 digit number.
So, the number of ways to select 5 digit out of 6 is ${}^6{C_5}$ which is equal to 6 $\left[ {\because {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}} \right]$
Now we have to arrange these numbers. So the possible arrangement is (5!).
So, the possible 5 digit numbers are $\left( {6 \times 5!} \right)$.
Now if the number is starting from 0 it is not a five digit number it is a four digit number so we have to subtract all four digit numbers from the possible five digit numbers.
So all the possible combination of five digits are (0, 1, 2, 3 and 4), (0, 1, 2, 3 and 5), (0, 1, 2, 4 and 5), (0, 1, 3, 4 and 5), (0, 2, 3, 4 and 5), (1, 2, 3, 4 and 5).
So out of these possible cases there are 5 possible cases in which numbers can start from 0.
So number of four digits numbers are $\left( {5 \times 4!} \right)$. (when the number is
starting from 0 the possible arrangements of the remaining digits is (4!))
So, the total 5 digit numbers are $ = \left[ {\left( {6 \times 5!} \right) - \left( {5 \times 4!}
\right)} \right] = \left[ {6 \times 120 - 5 \times 24} \right] = 720 - 120 = 600$
Now we have to find out the 5 digit numbers which are divisible by 6.
So we all know that if a number is divisible by 6 it is also divisible by 2 and 3.
So if a number is divisible by 3 the sum of all the digits should also be divisible by 3.
So the sum of all the digits is (0 + 1 + 2 + 3 + 4 + 5 = 15) so fifteen is divisible by 3.
Now we have to consider only 5 digits so there are two possible cases.
Case – 1 (0, 1, 2, 4, 5)
Case – 2 (1, 2, 3, 4, 5)
(Because if we subtract 0 or 3 the sum of digits is again divisible by 3 and when we subtract any other digit the sum of the digits is not divisible by 3)
Now consider case 1.
Case – 1 (0, 1, 2, 4, 5)
Now we have to find the numbers which are also divisible by 2.
As we know the number is divisible by 2 if the last digit of the number is even or 0.
So if the last digit is zero (0) so the number of possible ways which is divisible by 6 is = (4!).
$ \Rightarrow 4! = 4 \times 3 \times 2 \times 1 = 24$
Now if the last digit is not 0, so for this case there are two possible ways (2, 4)
And we all know numbers cannot start from 0. So, the first digit is chose from the remaining numbers
So, from (2, 4) one digit is placed as last digit and 0 cannot be the first digit so the remaining digits are (5 – 2 = 3)
So the number of ways to choose first digit is ${}^3{C_1} = 3$
So, first digit and last digit are chosen, now we have to make the arrangement for the middle three digits.
So, the possible number of arrangements is = (3!).
So, the numbers which are divisible by 6 and not starting from 0 are = $\left( {{}^3{C_1}
\times 3! \times 2} \right) = 3 \times 6 \times 2 = 36$
So, the total number of 5 digit number which is divisible by 6 for case 1 is (24 + 36 = 60)
Now consider case 2
Case -2 (1, 2, 3, 4, 5)
In this case there is no zero, so the last digit is filled by again in two possible ways (2, 4).
And the remaining four numbers can permute in (4!) ways.
So, the numbers which is divisible by 6 = $\left( {2 \times 4!} \right) = 48$
So, the total numbers including case 1 and case2 which are divisible by 6 = (60 +48 = 108).
So, the required probability (p) that the number is divisible by 6 is
\[ \Rightarrow p = \dfrac{{{\text{Favorable number of outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{{108}}{{600}} = \dfrac{{18}}{{100}} = 0.18\]
Now we have to convert this probability into percentage so we have to multiply by 100.
$ \Rightarrow p = 100\left( {0.18} \right) = 18$%.
So, this is the required answer.
Hence, option (c) is correct.
Note: Whenever we face such types of questions the key concept we have to remember is that the probability is the ratio of favorable number of outcomes to the total number of outcomes so, first find out the total number of outcomes as above then find out the favorable number of outcomes using the concept of divisibility rule of 6 which is stated above then divide these according to probability theorem and convert it into percentage by multiplying 100, we will get the required probability in percentage.
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