Question

How many numbers between 400 and 1000 can be formed with the digits 0, 2, 3, 4, 5, 6, if no digit is repeated in the same number?

Answer 1

In this question, first we will find the number of digits which can be placed at the hundredth place . From the given numbers only three numbers (4,5,6) can be placed at hundredth place. Now we will arrange digits at the two other places (tenth and ones place). This can be done in $^{5}\textrm{P}_{2}$ways.

Complete step-by-step answer:

In this question, number of digits given are: 0, 2, 3, 4, 5, 6,

And we have to make the number between 400 and 1000

Number between 400 and 1000 consist of three digits

First of all we see

Digit at the hundredth place which is greater than or equal to 4.

Hundredth place can be filled, by using the digits 4, 5 and 6 in 3 ways.

Now ten’s and unit’s places can be filled by the remaining 5 digits in $^{5}\textrm{P}_{2}$ ways.

Hence, the required number of numbers = 3 $\times ^{5}\textrm{P}_{2}$

= 3 $\times$ $\dfrac{{5!}}{{3!}}$

= 3 $\times$ $\dfrac{{5 \times 4 \times 3!}}{{3!}}$

= 3 $\times$ 20 =60

So, the required number of numbers is 60.

Note: We define the value of $0! = 1$ . Permutation is the arrangement of letters. The different groups or selection of number or object is a combination. Number of combinations is $^{n}\textrm{C}_{r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ where; the number of all combination of things, taken r times. $^{n}\textrm{C}_{n} = 1$ And $^{n}\textrm{C}_{0} = 0$ . A combination is an arrangement of items in which order does not matter but a permutation is an arrangement of items in a particular order which matters.