Question

The number of \[3\] digit numbers \[abc\] such that \[b < c\] is
A. \[450\]
B. \[405\]
C. \[400\]
D. \[410\]

Answer 1

Hint: We are asked to find the number of three digit numbers \[abc\] such that \[b < c\] . First find the total number of three digit numbers. Then find the number of ways for selecting the value of \[a\] using the combination formula. Find the number of three digit numbers with \[b = c\] . Using these values find the required answer.

Complete step-by-step answer:
Given to find the number of three digit numbers \[abc\] such that \[b < c\] .
We will use here the concept of combination. Using combination we can find the number of ways of selecting a number of items from a set of items. It is written as,
 \[^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] (i)
where \[^n{C_r}\] is the number of ways of selecting \[r\] number of items from a set of \[n\] number of items.
First we find the total number of three digit numbers. The smallest three digit number is \[100\] and the largest three digit number is \[999\] . So, the total number of three digit number will be,
 \[T = 900\]
 \[a\] can take the values \[1,2,3,4,5,6,7,8,9\] , we didn’t include \[0\] as putting \[a = 0\] would make the number a two digit number.
Therefore, set of values for a is \[\left\{ {1,2,3,4,5,6,7,8,9} \right\}\] . Using the formula from equation (i) the number of ways of selecting the value of \[a\] is,
 \[N(a){ = ^9}{C_1} = \dfrac{{9!}}{{1!\left( {9 - 8} \right)!}}\]
 \[ \Rightarrow N(a) = 9\]
The values for \[b\] and \[c\] can be \[\left\{ {0,1,2,3,4,5,6,7,8,9} \right\}\] .
We take the case when \[b = c\] . In this case the possible set of values for \[bc\] will be,
 \[\left\{ {00,11,22,33,44,55,66,77,88,99} \right\}\]
The number of ways of having \[b = c\] is,
 \[N(bc) = 10\]
Therefore, combining the above values we get the number ways of having a three digit number \[abc\] such that \[b = c\] is,
 \[N(abc) = N(a) \times N(bc)\]
Putting the values of \[N(a)\] and \[N(bc)\] we get,
 \[N(abc) = 9 \times 10 = 90\]
This means there \[90\] three digit numbers which have \[b = c\] .
The remaining three digit number will be,
 \[R = T - N(abc)\]
Putting the values of \[T\] and \[N(abc)\] we get,
 \[R = 900 - 90 = 810\]
From these remaining numbers half of the numbers will be such that \[b < c\] and half of the numbers will be such that \[b > c\] .
Calculating the half of the remaining numbers we get the required answer. That is, number of three digit number having the condition \[b < c\] is,
 \[N' = \dfrac{1}{2}R\]
Putting the value of \[R\] we get,
 \[N' = \dfrac{1}{2} \times 810 = 405\]
Therefore, the number of \[3\] digit numbers \[abc\] such that \[b < c\] is \[405\] .
Hence, the correct answer is option (B) \[405\] .
So, the correct answer is “Option B”.

Note: We have used here the concept of combination. There is no more term known as permutation. Sometimes students get confused between the two terms. Permutation means arranging a set of items in a sequence or order. The formula for permutation is \[P(n,r) = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] where \[P(n,r)\] is the number of ways of selecting and arranging \[r\] number of items from a set of \[n\] items.