GATE CSE 2010 | Question: 65

Question

Given digits $ 2, 2, 3, 3, 3, 4, 4, 4, 4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?

• A. $50$
• B. $51$
• C. $52$
• D. $54$

Comments

Notice -->

$\binom{9}{4}*4! - \binom{8}{3}*3!$ In this way you will get wrong answer because we are counting same number multiple times.

---

MY first approach was 7*8*7*6=2352 ways .
WHat is wrong in this apporach ? Am i not excluding the repeated parts?

---

We considered selections of ${\color{Red}{\textbf{k items}} }$ out of ${\color{Magenta}{\textbf{n possible options.}} }$

	$\textbf{With repetitions}$	$\textbf{Without repetitions}$
$\textbf{Ordred}$	$\text{Tuples}:n^{k}$	$k-\text{permutations}:\dfrac{n!}{(n-k)!}$
$\textbf{Unordered}$	$\text{Combinations with repetitions}:\binom{k + n - 1}{n-1} = \binom{k + n - 1}{k} $ Also known as Integer Equations - Stars and Bars	$\text{Combinations}:\binom{n}{k}$

Answer 1

For the case of the Number greater than 3000
Let us make the choice form where :-

1^st Position has 2 choices to make from (3 or 4)

2^nd Position has 3 choices to make form (2 or 3 or 4)

3^rd Position has 3 choices to make form (2 or 3 or 4)

4^th Position has 3 choices to make form (2 or 3 or 4)

 

Hence total of 2*3*3*3 = 54 Choices 
But here we have done a mistake as it can include number such as 4222, 3222 and 3333. So we need to subtract these. 

Hence net resultant will be = 54-3 = 51 Choices.