Question

Number of natural numbers upto one lakh, which contains $ 1,2,3, $ exactly once and remaining digits any time is –
A. $ 2940 $
B. $ 2850 $
C. $ 2775 $
D. $ 2680 $

Answer 1

Hint: Permutation is an ordered combination- an act of arranging the objects or numbers in the specific favourable order. Here, we will follow the basic concepts of three digits mathematical terms and apply the permutations for the given set of numbers.

Complete step-by-step answer:
One lakh as six digits, $ 100000 $
Smallest $ 3 $ digit number which contains $ 1,2,3, $ exactly once are arranged in
 $ 3! $ ways $ = 3 \times 2 \times 1 = 6 $ ways. ..... (A)
Now, four digit number which contains $ 1,2,3, $ exactly once and rest one digit can be occupied by $ 0,4,5,6,7,8,9 $ were
Case (I) $ 0 $ cannot be occupied on thousand’s place, $ 4,5,6,7,8,9 $
Can be arranged in $ {}^6{C_1}.4! $ ways $ = 6 \times 4 \times 3 \times 2 \times 1 = 144 $ .... (B)
Case (II) Here, $ 0 $ cannot be occupied on a thousand's place, but it can occupy any other place such as units, tens and hundreds.
It can be arranged in $ {}^3{C_1}.3! = 3 \times 3 \times 2 = 18 $ ways. .... (C)
Five digit number we are given that repetitions of other numbers except $ 1,2,3, $ can be done many times.
Case (I) $ 0 $ cannot be occupied on ten thousand’s place, $ 4,5,6,7,8,9 $
Can be arranged in $ {}^5{C_3}.3! \times 36 $ ways \[ = \dfrac{{5 \times 4 \times 3}}{{3 \times 2 \times 1}} \times 3 \times 2 \times 1 \times 36 = 2160\] .... (D)
Case (II) Here, $ 0 $ cannot be occupied on ten thousand’s place, but it can occupy any other place such as units, tens and hundreds, so with zero one time
It can be arranged in $ {}^4{C_1}.4! \times 6 = 4 \times 4 \times 3 \times 2 \times 6 = 576 $ ways. .... (E)
Case (III) Here $ 0 $ can be used twice and it can be arranged in
 $ {}^4{C_2}.3! = \dfrac{{4 \times 3}}{2} \times 3 \times 2 = 36 $ ...... (F)
Add the values of all the equations –
 $ \Rightarrow 6 + 144 + 18 + 2160 + 576 + 36 $
Simplify the above expression-
 $ 2940 $ ways.
So, the correct answer is “Option A”.

Note: Know the permutations and combinations concepts properly and apply accordingly. : In permutations, specific order and arrangement is the most important whereas a combination is used if the certain objects are to be arranged in such a way that the order of objects is not important.