How many odd positive integers with distinct digits can be created usi

Question

How many odd positive integers with distinct digits can be created using the digits 1, 2, 3, 4, 5 and 6?

A. 960
B. 975
C. 976
D. 978
E. 986

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PoojanB · Answer

1 digit number = 1,3,5 = 3 nos
2 digit number = _ _= for unit digit we have 3 ways to choose 1 no out of 3 odd nos (1,3,5)
 For tens digit we have 5 ways to choose 1 out of 5 remaining Nos .
 thus ,total Nos- 5*3=15
Likewise for 3 digit number =4*5*3=60
4 digit number = 3*4*5*3= 180
5 digit number= 2*3*4*5*3= 360
6 digit number= 1*2*3*4*5*3=360

total nos = 360+360+180+60+15+3
 = 978

Ans Choice -  D

lacktutor · Answer

How many odd positive integers with distinct digits can be created using the digits 1, 2, 3, 4, 5 and 6?

-----1 --->5!= 120 
-----3 --->5!= 120
-----5 --->5!= 120

----1 ---> 5*4*3*2= 120
----3 ---> 5*4*3*2= 120
----5 ---> 5*4*3*2= 120

---1 ---> 5*4*3=60
---3 ---> 5*4*3=60
---5 ---> 5*4*3=60

--1 ---> 5*4=20
--3 ---> 5*4=20
--5 ---> 5*4=20

-1 ---> 5
-3 ---> 5
-5 ---> 5

1 --->1
3 --->1
5 --->1

In total, 360+ 360+ 180+ 60+ 15+ 3= 978 
ANSWER (D).

JohnsonOoi · Answer

D

6digit odd integers = 360
5 digit odd integers = 360 
4 digit odd integers = 180
3 digit odd integers = 60
2 digit odd integers= 15
1 digit odd integers =3

PolarisTestPrep · Answer

Hello!

Here it is important to begin with a clear understanding of the question

By distinct digits, it is meant that the digits in the number do not repeat

For instance, 222,231 cannot be one of the possibilities because the number 2 repeats too often. Each of the digits in the ones, tens, hundreds, etc. position must be distinct from all the other digits.

We are given 6 different numbers 1, 2, 3, 4, 5, and 6 to construct with

We are told that the number must be positive, an integer, and odd

This means that the number can only end in 1, 3, or 5

Let's go methodically through the possibilities

How many 1 digit numbers can we make?

1, 3, and 5 = 3 different numbers that are positive odd integers

How many 2 digit numbers can we make?

We know that the ones value will be one of three numbers, either 1, 3, or 5

That means that for the tens position, 5 numbers will be left over to choose from and combine with the other number:

5 x 3 = 15 different possible combinations or distinct numbers

How many 3 digit numbers can we make?

After using a number in the ones positon and another number in the tens position, we are left with four numbers to choose from for the hundreds position

4 x 5 x 3 = 60 different possible combinations

How many 4 digit numbers can we make

There are 3 numbers to choose from left over for the thousands position

3 x 4 x 5 x 3 = 180 different possibilities

How many 5 digit numbers?

There are 2 numbers left to choose from to make sure that each digit is distinct from the others so we get:

2 x 3 x 4 x 5 x 3 = 360 different possibilities

And finally, how many 6 digit numbers can we make?

Well, there will only be 1 number left over to put in the hundred thousands position:

1 x 2 x 3 x 4 x 5 x 3 = 360 possibilities

Now we add up all the different numbers that can be made:

360 six digit numbers + 360 five digit numbers + 180 four digit numbers + 60 three digit numbers + 15 two digit numbers + 3 one digit numbers =

978 different numbers can be made

The answer is (D)

SaquibHGMATWhiz · Answer

Solution:

The most important thing to notice here is that the question is asking us " how many odd positive integers.... can be created "

Notice that the  question is not stating the number of digits of the interger. Therefore we have to consider all the possible scenarios

1 digit integer:

Only 3 possibilities: 1, 3 and 5

2 digit integers:

If we use the space-filling method, we will get   5  ×  3  = 15  
which is 3 choices (1, 3 and 5) for unit digit and 5 (after placing unit digit) choices for tens digit

3 digit integers:

If we use the space-filling method, we will get   5  ×  4  ×  3  = 60  
which is 3 choices (1, 3 and 5) for unit digit, 5 (after placing unit digit) choices for tens digit and 4 for the hundreds digit

4 digit integers:

If we use the space-filling method, we will get   5  ×  4  ×  3  ×  3  = 180

5 digit integers:

If we use the space-filling method, we will get   5  ×  4  ×  3  ×  2  ×  3  = 360

6 digit integers:

If we use the space-filling method, we will get   5  ×  4  ×  3  ×  2  ×  1  ×  3  = 360

Total odd integers possible  = 3+15+60+180+360+360=978

Hence the right answer is    Option D

Purnank · Answer

Given digits - 1 2 3 4 5 6
1 digit odd numbers - 3
2 digit odd numbers - 3*5 = 15
3 digit odd numbers - 3*5*4 = 60
...4 - 180, 5 - 360 and 6 - 360.
Add all = 3 + 15 + 60 + 180 + 360 + 360 = 978
  Answer is Option D.

benjamindg95 · Answer

A quick shortcut to prevent needing to dive deep into all calculations.

Focus on the digit units in the answer choices.

Once we have successfully calculated the choices for 1-digit numbers (3), and 2-digit numbers (15), we can recognize that all the remaining totals to calculate will end in 0 given they have a multiple of 10 (2 x 5). For example, 3 digits will give (5 x 4 x 3 = 60) options, 4 digits will give (5 x 4 x 3 x 3 = 180 options).

Knowing this, we can look to the answer choices to see (D) is the only option where the unit digit is 8, which we will need to be true for the 15 + 3 that is added to all these multiples of 10.

Hope this helps.

A_Nishith · Answer

1 digit integer:

Only 3 possibilities: 1, 3 and 5

2 digit integers:

If we use the space-filling method, we will get 5 × 3 = 15
which is 3 choices (1, 3 and 5) for unit digit and 5 (after placing unit digit) choices for tens digit

3 digit integers:

If we use the space-filling method, we will get 5 × 4 × 3 = 60
which is 3 choices (1, 3 and 5) for unit digit, 5 (after placing unit digit) choices for tens digit and 4 for the hundreds digit

4 digit integers:

If we use the space-filling method, we will get 5 × 4 × 3 × 3 = 180

5 digit integers:

If we use the space-filling method, we will get 5 × 4 × 3 × 2 × 3 = 360

6 digit integers:

If we use the space-filling method, we will get 5 × 4 × 3 × 2 × 1 × 3 = 360

Total odd integers possible =3+15+60+180+360+360=978

Hence the right answer is  Option D

How many odd positive integers with distinct digits can be created usi

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How many odd positive integers with distinct digits can be created usi

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