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How many 2 digit numbers can be formed using any of the 10 digits when

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chetan2u
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How many 2 digit numbers can be formed using any of the 10 digits when [#permalink] New post   17 Jan 2022, 06:24
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57% (00:47) correct 43% (00:54) wrong based on 28 sessions
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How many 2 digit numbers can be formed using any of the 10 digits when repetition is allowed?

(A) 50
(B) 70
(C) 90
(D) 100
(E) 180
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C
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chetan2u
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Re: How many 2 digit numbers can be formed using any of the 10 digits when [#permalink] New post   17 Jan 2022, 06:35
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chetan2u wrote:
How many 2 digit numbers can be formed using any of the 10 digits when repetition is allowed?
(A) 50
(B) 70
(C) 90
(D) 100
(E) 180



Mugdho

I have rewritten the query you had asked here, as that post is locked.

Two ways as you too mentioned

1) Permutation:
Two places AB
A can be any of the digits except 0, so 9.
B can be any of the 10 digits
So 9*10 or 90

2) Combination
You have taken it as 9C1*10C1*2!. That is wrong.
Permutation will be choose any two digits in 10C2 or 45 ways and they can be arranged in 2! ways, so 45*2=90 ways where digits are different.
Add ways when both digits are same, so 10 ways.
But subtract ways where 0 is in tens place, so 1*10 or 10 ways
Total 90+10-10 = 90 ways.
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Mugdho
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Re: How many 2 digit numbers can be formed using any of the 10 digits when [#permalink] New post   17 Jan 2022, 07:32
chetan2u wrote:
chetan2u wrote:
How many 2 digit numbers can be formed using any of the 10 digits when repetition is allowed?
(A) 50
(B) 70
(C) 90
(D) 100
(E) 180



Mugdho

I have rewritten the query you had asked here, as that post is locked.

Two ways as you too mentioned

1) Permutation:
Two places AB
A can be any of the digits except 0, so 9.
B can be any of the 10 digits
So 9*10 or 90

2) Combination
You have taken it as 9C1*10C1*2!. That is wrong.
Permutation will be choose any two digits in 10C2 or 45 ways and they can be arranged in 2! ways, so 45*2=90 ways where digits are different.
Add ways when both digits are same, so 10 ways.
But subtract ways where 0 is in tens place, so 1*10 or 10 ways
Total 90+10-10 = 90 ways.



chetan2u

but why do we have to take 2 digits at a time(10c2)?

There are 10 digits.
So for ten's digit we can choose any one from the 9 digits and for unit digit we can choose any one digit from the 10 digits. And then we then arrange them (as we do for other problems of this kind). Why this idea is not valid?!

Posted from my mobile device
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chetan2u
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Re: How many 2 digit numbers can be formed using any of the 10 digits when [#permalink] New post   18 Jan 2022, 05:30
Expert Reply
Mugdho wrote:
chetan2u wrote:
chetan2u wrote:
How many 2 digit numbers can be formed using any of the 10 digits when repetition is allowed?
(A) 50
(B) 70
(C) 90
(D) 100
(E) 180



Mugdho

I have rewritten the query you had asked here, as that post is locked.

Two ways as you too mentioned

1) Permutation:
Two places AB
A can be any of the digits except 0, so 9.
B can be any of the 10 digits
So 9*10 or 90

2) Combination
You have taken it as 9C1*10C1*2!. That is wrong.
Permutation will be choose any two digits in 10C2 or 45 ways and they can be arranged in 2! ways, so 45*2=90 ways where digits are different.
Add ways when both digits are same, so 10 ways.
But subtract ways where 0 is in tens place, so 1*10 or 10 ways
Total 90+10-10 = 90 ways.



chetan2u

but why do we have to take 2 digits at a time(10c2)?

There are 10 digits.
So for ten's digit we can choose any one from the 9 digits and for unit digit we can choose any one digit from the 10 digits. And then we then arrange them (as we do for other problems of this kind). Why this idea is not valid?!

Posted from my mobile device



Because that is exactly what combination means.
So 10 C2 is number of different combinations of pair of digits and then these pairs can be arranged in 2! within themselves.
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Re: How many 2 digit numbers can be formed using any of the 10 digits when [#permalink]
18 Jan 2022, 05:30
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