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How many odd numbers with 4 different digits, can be formed using the

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Bunuel
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How many odd numbers with 4 different digits, can be formed using the [#permalink] New post   14 Jun 2019, 00:13
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How many odd numbers with 4 different digits, can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8? (No repetition is allowed)

A. 71

B. 200

C. 210

D. 840

E.1680
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How many odd numbers with 4 different digits, can be formed using the [#permalink] New post   14 Jun 2019, 00:19
Bunuel wrote:
How many odd numbers with 4 different digits, can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8? (No repetition is allowed)

A. 71

B. 200

C. 210

D. 840

E.1680



Let the odd digit be ABCD

Number of options for D = 4 (1,2,3 or 4)
A = 7, B = 6, C = 5
Total number of options = 7*6*5*4 = 840

IMO Option D
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Re: How many odd numbers with 4 different digits, can be formed using the [#permalink] New post   14 Jun 2019, 00:19
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Bunuel wrote:
How many odd numbers with 4 different digits, can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8? (No repetition is allowed)

A. 71

B. 200

C. 210

D. 840

E.1680



Let us first take the units digit as there is a restriction there..
If the number is odd, the units digit can be any of 1, 3, 5 or 7 --- so 4 ways
The rest three digits can be in 7*6*5 ways..

Total ways --- ABCD = 7*6*5*4=840.... ABC in 7*6*5 ways and D in 4 ways

D
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Re: How many odd numbers with 4 different digits, can be formed using the [#permalink] New post   14 Jun 2019, 00:25
Bunuel wrote:
How many odd numbers with 4 different digits, can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8? (No repetition is allowed)

A. 71

B. 200

C. 210

D. 840

E.1680


total odd= 1,3,5,7 ;4
so last place we have 4 option
for rest 3 positions ; 7*6*5*4 = 840
IMO D
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Re: How many odd numbers with 4 different digits, can be formed using the [#permalink] New post   08 Jul 2019, 04:10
The rest three digits can be in 7*6*5 ways..

how are you deciding to multiply by 4 ? pls help me to understand better...

chetan2u wrote:
Bunuel wrote:
How many odd numbers with 4 different digits, can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8? (No repetition is allowed)

A. 71

B. 200

C. 210

D. 840

E.1680



Let us first take the units digit as there is a restriction there..
If the number is odd, the units digit can be any of 1, 3, 5 or 7 --- so 4 ways
The rest three digits can be in 7*6*5 ways..

Total ways --- ABCD = 7*6*5*4=840.... ABC in 7*6*5 ways and D in 4 ways

D
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Re: How many odd numbers with 4 different digits, can be formed using the [#permalink] New post   18 Sep 2021, 00:33
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Correct option : D - 840

Given: {1,2,3,4,5,6,7,8}
Condition: No repetition of numbers
To Form - Number of ways - 4 digit Odd number can be formed: (Not divisiable by 2)
Number at unit number {1,3,5,7} are only possiable to form Odd Number

A - 1000th place
B - 100th place
C - 10th place
D - unit place

D - Unit place option {1,3,5,7} - 4 option (without repetition)
C - 10th place - 7 ways can be arranged
B - 100th place - 6 ways can be arranged
A - 1000th place - 5 ways can be arranged

4 x 7 x 6 x 5 = 840

Additional information : if it was with repetition :
solution will be : 8 x 8 x 8 x 8 = 4096 for 4 digit number of ways to be arranged
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Re: How many odd numbers with 4 different digits, can be formed using the [#permalink]
18 Sep 2021, 00:33
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