Out of the six digits 1, 2, 3, 4, 5 and 6; how many two digit numbers

Out of the six digits 1, 2, 3, 4, 5 and 6; how many two digit numbers can be formed without repetition of digits ?

[amp_mcq option1=”6″ option2=”15″ option3=”30″ option4=”40″ correct=”option3″]

This question was previously asked in
UPSC CAPF – 2024
30 two-digit numbers can be formed without repetition of digits.
The given digits are 1, 2, 3, 4, 5, and 6. There are a total of 6 distinct digits.
We need to form two-digit numbers without repetition. A two-digit number has a tens digit and a units digit.
For the tens digit, we can choose any of the 6 available digits. So, there are 6 choices for the tens digit.
Since repetition of digits is not allowed, the units digit must be different from the digit chosen for the tens place. After choosing one digit for the tens place, there are 5 digits remaining. So, there are 5 choices for the units digit.
The total number of two-digit numbers that can be formed is the product of the number of choices for each position:
Total number of numbers = (Number of choices for tens digit) $\times$ (Number of choices for units digit)
Total number = $6 \times 5 = 30$.
This problem is a case of permutation, specifically finding the number of permutations of 6 distinct items taken 2 at a time, denoted as P(6, 2) or $_6P_2$. The formula for permutations is $P(n, k) = \frac{n!}{(n-k)!}$, where n is the total number of items and k is the number of items to choose.
Here, n = 6 (the digits) and k = 2 (the number of digits in the number).
$P(6, 2) = \frac{6!}{(6-2)!} = \frac{6!}{4!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} = 6 \times 5 = 30$.