Permutations and Combinations

Question

 (i) Find how many different numbers can be formed using 4 of the digits 1, 2, 3, 4, 5, 6 and 7 if no digit is repeated.

(iii) odd and less than 3000

 

 

 

 

ans

480

Answer 1

(i) We have 7 digits. We take 4 of them to form a number.

For first place we can take any of them, thus number of choices is 7.

For second place we are to choose another digit, therefore number of choices is less by one and equal 6.

For third one we can choose out of 5 digits which are not the ones we picked up already. 

And for the last position we are left with 4 digits. 

7*6*5*4=840

(iii) Odd and less than 3000 mean the following restrictions on our digits:

•  The last one may be only 1, 3, 5, 7.
•  The first one may be only 1 or 2. 

We can split the resulting numbers into 2 categories:  

a) starting with 1

b) starting with 2 

For a) we have 1 choice for first place, 6 for second, 5 for third and... it is difficult to say about the last one because one of 

the odd numbers could be chosen already and we cannot know what was it. 

So let us change the order in which we pick up numbers. We will start with the last one. For this we have only numbers 3, 5 and 7, because 1 will be the first. Hence we have 3 choices. Then we have 5 for the third one (because 1 should be exluded together with the one which is now in the last place) , and we have 4 choices for second place, and 1 for the first.

3*5*4*1=60.

For b) we start to fill numbers from the last position as well, keeping in mind that 2 is already taken for the first place. Thus we have 4 odd numbers for the last position, 5 for the third, 4 for second and  1 for the first.

4*5*4*1=80.

Altogether: 60+80=140.