Question

The number of five – digit numbers that contain 7 exactly once is:
(a) $\left( 41 \right)\left( {{9}^{3}} \right)$
(b) $\left( 37 \right)\left( {{9}^{3}} \right)$
(c) $\left( 7 \right)\left( {{9}^{4}} \right)$
(d) $\left( 41 \right)\left( {{9}^{4}} \right)$

Answer 1

Hint: First of all, consider 2 cases when 7 is at the leftmost place and when 7 is not at the leftmost place. Now, count the total number in the first and second cases by filling the remaining places with the remaining numbers and add these numbers in two cases to get the final answer.

Complete step-by-step answer:
In this question, we have to find the number of five-digit numbers that contain 7 exactly once. Let us consider five places for 5 digit numbers.
In this question, we are given that 7 should come exactly once, so let us consider two cases: when 7 is at the leftmost or 10000th place and when 7 is at any other place except 10000th place. Let us consider each case individually.
Case 1: When 7 is at the leftmost or 10000th place.
$\underline{7}\times \underline{9}\times \underline{9}\times \underline{9}\times \underline{9}$
We know that from 0 – 9, we have a total of 10 numbers out of which we have already used 1 number that is 7. So, we can fill the next place that is 1000th place with any one of the remaining 9 numbers. Similarly, we can again fill the next place with any one of the 9 numbers. We can fill the remaining two in a similar way.
Hence, the possible numbers that can be formed in this case
$=9\times 9\times 9\times 9$
$={{9}^{4}}$
Case 2: When 7 is not at leftmost or 10000th place

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In this case, we have 4 possibilities for the number 7 that is 4 places to fill the number 7 except the 10000th place. Now, the leftmost place or 1000th place can be filled in only 8 ways that are from 1 – 9 except 7. We can’t fill 0 in the leftmost place because then it won’t be a five-digit number. Now, we can fill the remaining 3 places in 9 ways that are from 0 – 9 except 7.
So, the total number that can be formed in this case
$=4\times 8\times {{\left( 9 \right)}^{3}}$
By adding the numbers formed in two cases, we get the total numbers as
$={{9}^{4}}+\left( 4\times 8\times {{9}^{3}} \right)$
$={{9}^{3}}\left[ 9+{{3}^{2}} \right]$
$={{9}^{3}}.41$
Hence, we get the total number of five-digit numbers that contain 7 exactly once is $\left( 41 \right)\left( {{9}^{3}} \right)$.
So, option (a) is the right answer.

Note: In case 2 of the above question, students often fill the leftmost place with 9 numbers as well, that is they forget to exclude 0. So, this must be taken care of because 0 at the leftmost place of a number is meaningless. Also, in the above question, some students mark the option (b) that is $\left( 41 \right)\left( {{9}^{4}} \right)$ in a hurry. So, this mistake must be avoided by looking at the options properly and not just in this question, but in every question.