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Question

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a. 2940

b. 2850

c. 2775

d. 2680

Answer
Verified

Hint: Here 1, 2, 3 places are exactly 1. Take care of 3, 4 and 5 digits, with and without zero for 4 digits. Without zero and repetition, with 1 and 2 zero’s for 5 digit numbers. Find the values and add them.

Complete step-by-step answer:

The natural number up to 1 lakh is considered, which means that we have 100000 numbers. The total number of digits is 6. We can’t consider a number greater than 5 digit because it might be greater than the 6 digit number given to us.

We have been given that it contains 1, 2, 3 exactly ones.

If we are taking the first 3 digit number we can arrange it in 3! ways.

\[3\times 2\times 1=3!=6\] ways – (1)

Now let us take a 4 digit number. Consider numbers from 0 – 9, apart from 1, 2, 3 the other digits present are 0, 4, 5, 6, 7, 8, 9 i.e. total of 7 digits. Zero can’t come in the place of x.

So let us find a 4 digit number without any zero’s.

x 1 2 3 , the places of 1, 2 and 3 are already fixed and for the place x any number from (4, 5, 6, 7, 8, 9) can come.

i.e. \[{}^{6}{{C}_{1}}\] ways are possible to arrange digits in place of x.

\[{}^{6}{{C}_{1}}\] is in the form of \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].

\[{}^{6}{{C}_{1}}=\dfrac{6!}{1!\left( 6-1 \right)!}=\dfrac{6!}{5!\times 1!}=\dfrac{6\times 5!}{5!1!}=6\]

Thus the 4 digit number without any zero, can be arranged \[{}^{6}{{C}_{1}}\times 4!\] ways. – (2)

Now let us find a way in which a 4 digit number forms with zero.

As zero can’t come in place of x i.e., x _ _ _

But zero can come in the rest of 3 places in \[{}^{3}{{C}_{1}}\] ways and the other numbers can be arranged in 3! ways.

\[\therefore \] 4 digit numbers with zero can be arranged as \[{}^{3}{{C}_{1}}\times 3!\] ways. – (3)

Now let us arrange a 5 digit number now. In this case repetition is allowed, _ _ _ _ _. Let us consider the case without zero. In these 5 places 1, 2, 3 can be arranged in any place. Thus we can arrange it in \[{}^{5}{{C}_{3}}\] ways. And in these 3 places 1, 2, 3 can be arranged in any way in 3! ways.

Thus the arrangement of 1, 2, 3 becomes \[{}^{5}{{C}_{3}}\times 3!\] ways.

Now there are 2 places left to fill and we can fill any digits from (4, 5, 6, 7, 8, 9). Thus as repetition is allowed both the 2 places can be arranged in 6 ways. Thus it becomes \[6\times 6=36\].

Thus the total arrangement of 5 digit number without zero and with repetition is

\[{}^{5}{{C}_{3}}\times 3!\times 36\] ways. – (4)

Now let us find the case of a 5 digit number with zero. Apart from the 3 places fixed we have 2 places. Zeroes can come in one place or in both places.

Thus let us first find the case with one zero in the 5 digit number.

Now there are 5 places, x _ _ _ _.

Zero can’t come in the place marked x. But it can come anywhere in the next 4 places. So it becomes \[{}^{4}{{C}_{1}}\]. Now after arranging zero, there are still 4 places left. So 3 places are occupied by (1, 2, 3) and the \[{{4}^{th}}\] place can be occupied by any digit from (4, 5, 6, 7, 8, 9).

Thus the total way to select 5 digit with one zero \[={}^{4}{{C}_{1}}\times 4!\times 6\] - (5)

Now let us go to the last case where the 5 digit number is with 2 zeroes, x _ _ _ _. Zero can’t come in the place marked x. But the 2 zero’s can come in any other place. So \[{}^{4}{{C}_{2}}\] ways and in the rest 3 places comes (1, 2, 3) which is arranged in 3! ways.

\[\therefore \] Total ways to select 5 digit with two zero \[={}^{4}{{C}_{2}}\times 3!\] ways – (6)

Thus we have seen a lot of cases with 3 digit, 4 digit and 5 digit numbers. Let us now add all these cases,

= three digit number + 4 digit number without zero + 4 digit number with zero + 5 digit without zero + 5 digit with 1 zero + 5 digit with 2 zero.

\[\begin{align}

& =6+{}^{6}{{C}_{1}}\times 4!+{}^{3}{{C}_{1}}\times 3!+{}^{5}{{C}_{3}}\times 3!\times 36+{}^{4}{{C}_{1}}\times 4!\times 6+{}^{4}{{C}_{2}}\times 3! \\

& =6+\left( 6\times 4\times 3\times 2\times 1 \right)+\left( 3\times 3\times 2\times 1 \right)+\left( 10\times 3\times 2\times 36 \right)+\left( 4\times 4\times 3\times 2\times 6 \right)+\left( 6\times 3\times 2 \right) \\

& \because {}^{5}{{C}_{3}}=\dfrac{5!}{3!\left( 5-3 \right)!}=\dfrac{5!}{2!3!}=\dfrac{5\times 4\times 3!}{2\times 3!}=\dfrac{5\times 4}{2}=5\times 2=10 \\

& \because {}^{4}{{C}_{2}}=\dfrac{4!}{2!\left( 4-2 \right)!}=\dfrac{4!}{2!2!}=\dfrac{4\times 3\times 2}{2\times 2}=6 \\

& =6+144+18+2160+576+36 \\

& =2940 \\

\end{align}\]

Thus we got the number of natural numbers as 2940.

\[\therefore \] Option (a) is the correct answer.

Note: Use imagination of how to fill the places. 1, 2, 3 numbers are common for this number and these 3 digits can come anywhere in the number. So don’t fix these numbers. Remember in case of without zero, as numbers are there without zeroes.

Complete step-by-step answer:

The natural number up to 1 lakh is considered, which means that we have 100000 numbers. The total number of digits is 6. We can’t consider a number greater than 5 digit because it might be greater than the 6 digit number given to us.

We have been given that it contains 1, 2, 3 exactly ones.

If we are taking the first 3 digit number we can arrange it in 3! ways.

\[3\times 2\times 1=3!=6\] ways – (1)

Now let us take a 4 digit number. Consider numbers from 0 – 9, apart from 1, 2, 3 the other digits present are 0, 4, 5, 6, 7, 8, 9 i.e. total of 7 digits. Zero can’t come in the place of x.

So let us find a 4 digit number without any zero’s.

x 1 2 3 , the places of 1, 2 and 3 are already fixed and for the place x any number from (4, 5, 6, 7, 8, 9) can come.

i.e. \[{}^{6}{{C}_{1}}\] ways are possible to arrange digits in place of x.

\[{}^{6}{{C}_{1}}\] is in the form of \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].

\[{}^{6}{{C}_{1}}=\dfrac{6!}{1!\left( 6-1 \right)!}=\dfrac{6!}{5!\times 1!}=\dfrac{6\times 5!}{5!1!}=6\]

Thus the 4 digit number without any zero, can be arranged \[{}^{6}{{C}_{1}}\times 4!\] ways. – (2)

Now let us find a way in which a 4 digit number forms with zero.

As zero can’t come in place of x i.e., x _ _ _

But zero can come in the rest of 3 places in \[{}^{3}{{C}_{1}}\] ways and the other numbers can be arranged in 3! ways.

\[\therefore \] 4 digit numbers with zero can be arranged as \[{}^{3}{{C}_{1}}\times 3!\] ways. – (3)

Now let us arrange a 5 digit number now. In this case repetition is allowed, _ _ _ _ _. Let us consider the case without zero. In these 5 places 1, 2, 3 can be arranged in any place. Thus we can arrange it in \[{}^{5}{{C}_{3}}\] ways. And in these 3 places 1, 2, 3 can be arranged in any way in 3! ways.

Thus the arrangement of 1, 2, 3 becomes \[{}^{5}{{C}_{3}}\times 3!\] ways.

Now there are 2 places left to fill and we can fill any digits from (4, 5, 6, 7, 8, 9). Thus as repetition is allowed both the 2 places can be arranged in 6 ways. Thus it becomes \[6\times 6=36\].

Thus the total arrangement of 5 digit number without zero and with repetition is

\[{}^{5}{{C}_{3}}\times 3!\times 36\] ways. – (4)

Now let us find the case of a 5 digit number with zero. Apart from the 3 places fixed we have 2 places. Zeroes can come in one place or in both places.

Thus let us first find the case with one zero in the 5 digit number.

Now there are 5 places, x _ _ _ _.

Zero can’t come in the place marked x. But it can come anywhere in the next 4 places. So it becomes \[{}^{4}{{C}_{1}}\]. Now after arranging zero, there are still 4 places left. So 3 places are occupied by (1, 2, 3) and the \[{{4}^{th}}\] place can be occupied by any digit from (4, 5, 6, 7, 8, 9).

Thus the total way to select 5 digit with one zero \[={}^{4}{{C}_{1}}\times 4!\times 6\] - (5)

Now let us go to the last case where the 5 digit number is with 2 zeroes, x _ _ _ _. Zero can’t come in the place marked x. But the 2 zero’s can come in any other place. So \[{}^{4}{{C}_{2}}\] ways and in the rest 3 places comes (1, 2, 3) which is arranged in 3! ways.

\[\therefore \] Total ways to select 5 digit with two zero \[={}^{4}{{C}_{2}}\times 3!\] ways – (6)

Thus we have seen a lot of cases with 3 digit, 4 digit and 5 digit numbers. Let us now add all these cases,

= three digit number + 4 digit number without zero + 4 digit number with zero + 5 digit without zero + 5 digit with 1 zero + 5 digit with 2 zero.

\[\begin{align}

& =6+{}^{6}{{C}_{1}}\times 4!+{}^{3}{{C}_{1}}\times 3!+{}^{5}{{C}_{3}}\times 3!\times 36+{}^{4}{{C}_{1}}\times 4!\times 6+{}^{4}{{C}_{2}}\times 3! \\

& =6+\left( 6\times 4\times 3\times 2\times 1 \right)+\left( 3\times 3\times 2\times 1 \right)+\left( 10\times 3\times 2\times 36 \right)+\left( 4\times 4\times 3\times 2\times 6 \right)+\left( 6\times 3\times 2 \right) \\

& \because {}^{5}{{C}_{3}}=\dfrac{5!}{3!\left( 5-3 \right)!}=\dfrac{5!}{2!3!}=\dfrac{5\times 4\times 3!}{2\times 3!}=\dfrac{5\times 4}{2}=5\times 2=10 \\

& \because {}^{4}{{C}_{2}}=\dfrac{4!}{2!\left( 4-2 \right)!}=\dfrac{4!}{2!2!}=\dfrac{4\times 3\times 2}{2\times 2}=6 \\

& =6+144+18+2160+576+36 \\

& =2940 \\

\end{align}\]

Thus we got the number of natural numbers as 2940.

\[\therefore \] Option (a) is the correct answer.

Note: Use imagination of how to fill the places. 1, 2, 3 numbers are common for this number and these 3 digits can come anywhere in the number. So don’t fix these numbers. Remember in case of without zero, as numbers are there without zeroes.

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