Answer to Question #241337 in Statistics and Probability for Nghelo

Question

Supposed repetition are not allowed.
a) How many three-digit numbers can be formed from the six digits 2, 3, 5, 6, 7 and 9?
b) How many are less than 500? [2]
c) How many are even? [2]
3.2 Simplify
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Answer 1

a.

To find the required solution, the concept of permutations is applied. It is given by;

${}^n{P}_k=\frac{n!}{(n-k)!}$

n= total number of elements in a set.

k= number of selected elements arranged in a specific order

To form three-digit numbers from the digits 2,3,5,6,7 and 9, there are n=6 digits and k=3.

Therefore,

${}^6{P}_3=\frac{6!}{(6-3)!}=\frac{6!}{3!}=6*5*4=120$

The number of three-digit numbers which can be formed from the digits 2,3,5,6,7 and 9 is 120

b.

For the numbers formed to be less than 500, the 100th's place must either be 2 or 3. This is described in the cases below;

Case 1: 2 is in the 100th's place.

They have already put the digit 2 in the 100th's place. There remains 5 distinct digits to occupy the 10th's place. In the unit's place, 4 distinct digits may occupy this place apart from 2 and the digit in the 10th's place. Therefore, total number of combinations for this case =1 * 5 * 4=20

Case 2: 3 is in the 100th's place.

The digit 3 is already in the 100th's place hence there are 5 distinct digits that may occupy the 10th's place and 4 distinct digits to occupy the unit's place apart from 3 and the digit in the 10th's place. Thus, total number of combinations=1 * 5 * 4=20.

Total number of combinations without repeating the digits =20+20=40.

So, the total number of three-digit numbers formed that are less than 500 is 40.

c.

For the numbers formed by the digits 2,3,5,6,7 and 9 to be even, the unit's digit must be 2 or 6.This can be expressed in two cases as below;

Case 1; numbers ending with 2.

Since 2 is already in the unit's place, some other digit should occupy the 10th's place. There are 5 other digits which can occupy this place.

Let's come to 100th's place. Apart from 2 and the digit that is already put in the 10th's place, there are 4 distinct digits which may now occupy the 100th's place.

Thus, total number of combinations = 4 * 5 * 1 = 20

Case 2: Numbers ending with 6.

The digit 6 is already in the unit's place. Of the remaining 5 distinct digits, another digit should occupy the 10th's place. In the 100th's place, 4 distinct digits may occupy this place apart from 6 and the digit in the 10th's place.

Therefore, total number of combinations for this case =4 * 5 * 1=20

Thus, total number of even numbers formed without the digits being repeated is 20+20=40

There are 40 even numbers formed.