Question

How many four-digit numbers can be formed by using each of the digits 9. 2, 0, 4 only once? (1) 24 (2) 18 (3) 20 (4) 12​

Answer 1

Answer:

P_Total there are 6 digits, from which we have to form 4 digits number.

We need to choose 4 digits randomly and form a number without repeating any digit.

In this process, we need to determine the number of possibilities where order does not matter.

So, we use the permutation formula and that is

n

P

r

=

(n−r)!

n!

, where n is the total number of objects and r is the number of objects taken at a time.

Given, n=6 and r=4

∴

6

P

4

=

(6−4)!

6!

=

2!

6!

=

2!

6×5×4×3×2!

=6×5×4×3

=360

∴

6

P

4

=360

∴ In 360 ways the 4-digits numbers can be form using the digits 1,2,3,7,8,9.

From these, the even numbers must contain the last digit as 2 or 8.

If we fix 2 or 8 at 4

th

place, then there are

5

P

3

ways for each.

∴ Total number of even numbers =2×

5

P

3

=2×

(5−3)!

5!

=2×

2!

5!

=2×

2×1

5×4×3×2×1

=120

∴ there are total 120 even numbers.

Answer 2

18, 4-digit numbers can be formed by using 9,2,0,4 only once.

Option '2' is correct.

Given:

• Four digits: 9,2,0,4

To find:

• How many four-digit numbers can be formed by using each of the digits only once ?
• (1) 24
• (2) 18
• (3) 20
• (4) 12

Solution:

Concept to be used:

When 'r' numbers are arranging from 'n' numbers without repeition

• The total number of ways are $\bf ^nP_r = \frac{n!}{(n - r)!}$
• 0!=1

Step 1:

Write the possible combination for the first digit.

9 _ _ _

2 _ _ _

4 _ _ _

'0' cannot be the first digit, because the number than formed will be a 3-digit number.

Step 2:

Find the total numbers.

For the first arrangement, 9 _ _ _

Three numbers are left, i.e. 2,0 and 4.

Possible number of arrangements of last three numbers:$^3P_3= \frac{3 \times 2 \times 1}{1}$

$^3P_3 = 6$

Thus,

By taking 9 as first digit, total 6 four digit numbers can be formed.

Similarly,

For the second and third arrangements; 6-6 numbers can also be formed.

Total four digit numbers without repeating the digits are $= 6 + 6 + 6$

$= 18$

Thus,

18, 4-digit numbers can be formed by using 9,2,0,4 only once.

Option '2' is correct.

Learn more:

1) how many 3 digit numbers can be formed from the digits 2 3 5 8 9 without repetition which are exactly divisible by 4

https://brainly.in/question/21389212

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