Question

Consider digits 1, 2, 3, 4, 5, 6 and 7. Using these digits, numbers of five digits are formed. Then, the probability of these five digit numbers that have odd digits at their both ends only, is (repetition not allowed).
$
  \left( A \right)\dfrac{3}{7} \\
  \left( B \right)\dfrac{1}{7} \\
  \left( C \right)\dfrac{2}{7} \\
  \left( D \right){\text{ None of these}} \\
$

Answer 1

Hint: In this question we have to form a 5 digit number using the given number, repetition is not allowed means we can’t choose a digit twice while forming a particular number. Firstly calculate the total 5 digit numbers that can be formed out of the given 7 digits using the concept of permutation and combination. Now in order to compute the probability that the numbers formed have odd digits at their both ends, form all the 5 digits numbers using the given number that have odd digits at both the ends. Apply the basic probability formula to get the answer.

Complete Step-by-Step solution:
Given digits (1, 2, 3, 4, 5, 6 and 7).
Now first find out the total numbers of five digits.
So the number of ways to choose 5 digits out of 7 are $ = {}^7{C_5}$.
Now we have to arrange these 5 digits so the number of ways to arrange them are (5!).
So the total number of five digits is $ = {}^7{C_5} \times 5!$.
Now we have to find the five digit numbers that have odd digits at their both ends only.
So out of seven digits the number of odd digits are (1, 3, 5 and 7) so there are four odd digits.
So the end digits of the five digit numbers is filled by ${}^4{C_2}$ number of ways .
And the arrangements of end digits of the five digit number are (2!).
So the end digits of the five digit numbers are filled by the odd digits and the number of ways is
$ = 2! \times {}^4{C_2}$
Now the remaining three middle digits of the five digits numbers are filled by remaining numbers (i.e. two odd digits are placed at both ends so the remaining digits are (7 – 2) = 5)
So the number of ways to fill middle digits of a five digit numbers are ${}^5{C_3}$
And the number of arrangements of these numbers are (3!)
So the total number of ways to fill middle digits of a five digit number are ${}^5{C_3} \times 3!$
So the possible number of ways of the five digit numbers having odd digits at their both ends only are
$ \Rightarrow 2! \times {}^4{C_2} \times {}^5{C_3} \times 3!$
Now as we now that the probability is the ratio of favorable number of outcomes to the total number of outcomes so the required probability (P) is,
$ \Rightarrow P = \dfrac{{2! \times {}^4{C_2} \times {}^5{C_3} \times 3!}}{{{}^7{C_5} \times 5!}}$
Now simplify the above equation we have,
\[ \Rightarrow P = \dfrac{{2 \times \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} \times \dfrac{{5!}}{{3!\left( {5 - 3} \right)!}} \times 6}}{{\dfrac{{7!}}{{5!\left( {7 - 5} \right)!}} \times 120}}\] $\left[ {\because {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}},5! = 120,3! = 6,2! = 2} \right]$
Now again simplify the above equation we have,
\[ \Rightarrow P = \dfrac{{2 \times 6 \times 10 \times 6 \times 2}}{{7 \times 6 \times 120}} = \dfrac{{10}}{{35}} = \dfrac{2}{7}\]
So this is the required probability.
Hence option (C) is correct.

Note: Whenever we face such type of problems the key concept is to have the good understanding of the information being provided in the question statement, the key point here is never forget to use permutation concept after choosing these 5 digits from 7 digits as the numbers can also rearrange amongst themselves and that needs to be counted as well. This concept along with a basic probability formula will help get on the right track to get the answer.