A seven digit number is formed using digit 3,3,4,4,4,5,5. The probability, that number so formed is

Q: Wait, does it matter which '4' I put at the end? They look the same!

That is a brilliant observation! Because the '4's are identical, it doesn't matter *which* specific '4' you place at the end. Mathematically, we treat them as indistinguishable, which is why we use the division by $3!$ in the total count and $2!$ in the favorable count to account for this symmetry.

Q: Why do we divide by $2!$ and $3!$ in the denominator?

Think of it as 'correcting' for overcounting. If all digits were distinct, we would have $7!$ arrangements. But since swapping two identical '3's doesn't create a new number, we divide by $2!$ to remove those redundant counts. We do this for every repeating digit group to ensure we only count unique arrangements.

Q: Can I solve this using the ratio of the number of 4s to the total number of digits?

That is a very intuitive shortcut! Since the probability of any specific digit occupying the last position is proportional to its frequency in the set, the probability of the last digit being a '4' is simply $\frac{\text{count of 4s}}{\text{total digits}} = \frac{3}{7}$. This is a powerful way to verify your result!

Question

A seven digit number is formed using digit 3,3,4,4,4,5,5. The probability, that number so formed is divisible by 2, is :

Options:

(A) $\frac{6}{7}$

(B) $\frac{4}{7}$

(C) $\frac{3}{7}$

(D) $\frac{1}{7}$

Sigma Percentile · Accepted Answer

### Analyzing the DNA of the Number

Before we jump into calculations, let us look at the raw material. We have seven digits: two threes, three fours, and two fives.

Our goal is to form a seven-digit number. If all these digits were distinct, the total number of arrangements would simply be $7!$.

However, they are not distinct. We must account for the fact that swapping two identical digits does not create a new number using the permutation formula for multisets.

### The Total Universe of Arrangements

To find the total number of possible arrangements, $n(S)$, we use the formula for permutations of a multiset:

$$n(S) = \frac{7!}{2! \cdot 3! \cdot 2!}$$

Here, $7!$ is the total number of permutations if all digits were distinct. We divide by $2!$ for the two threes, $3!$ for the three fours, and $2!$ for the two fives.

Calculating this, we get:

$$n(S) = \frac{5040}{2 	imes 6 	imes 2} = \frac{5040}{24} = 210$$

This is our sample space—the total number of unique seven-digit numbers we can form.

### The Constraint of Divisibility

Now, we consider the condition: the number must be divisible by $2$. In our base-10 system, a number is divisible by $2$ if and only if its last digit is even.

Looking at our set $\{3, 3, 4, 4, 4, 5, 5\}$, the only even digit available is $4$. Therefore, for our number to be even, the last digit must be a $4$.

We fix one $4$ at the unit place to satisfy this constraint.

### The Favorable Universe

With one $4$ fixed at the end, we are left with six slots to fill and six digits remaining: $\{3, 3, 4, 4, 5, 5\}$.

Notice that we now have two threes, two fours, and two fives. The number of ways to arrange these remaining digits, $n(E)$, is given by:

$$n(E) = \frac{6!}{2! \cdot 2! \cdot 2!}$$

Calculating this, we get:

$$n(E) = \frac{720}{2 	imes 2 	imes 2} = \frac{720}{8} = 90$$

These are our favorable outcomes.

### The Final Synthesis

We have our total outcomes ($210$) and our favorable outcomes ($90$). The probability $P(E)$ is the ratio of favorable outcomes to total outcomes:

$$P(E) = \frac{n(E)}{n(S)} = \frac{90}{210}$$

Simplifying this fraction, we cancel the zeros to get $\frac{9}{21}$. Dividing both the numerator and denominator by $3$, we arrive at the elegant result:

**$$P(E) = \frac{3}{7}$$**

This is the probability that our number is divisible by $2$. By breaking down the problem into the total universe and the constrained universe, we have navigated the complexity of permutations with ease.

Animated Solution for Mathematics - Probability: A seven digit number is formed using digit 3,3,4,4,4,5,5. The probability, that number so formed is divisible by 2, is :

Select Answer:

Visualized Solution

Analyze Given Digits

Total Arrangements Setup

Total Arrangements Calculation

Condition for Divisibility by $2$

Fixing the Last Digit

Favorable Arrangements Setup

Favorable Arrangements Calculation

Final Probability Calculation

Conclusion and Summary

The Sigma Insight: Classical Definition of Probability

Analyzing the DNA of the Number

The Total Universe of Arrangements

The Constraint of Divisibility

The Favorable Universe

The Final Synthesis

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