Quantitative Aptitude and Reasoning Archives

121. Which one of the following fractions is greater than $\frac{2}{7}$ and

Which one of the following fractions is greater than $\frac{2}{7}$ and less than $\frac{7}{9}$?

[amp_mcq option1=”$\frac{15}{19}$” option2=”$\frac{11}{17}$” option3=”$\frac{11}{14}$” option4=”$\frac{17}{21}$” correct=”option2″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

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To find the fraction greater than $\frac{2}{7}$ and less than $\frac{7}{9}$, we can compare each option with the given bounds. We can convert the fractions to decimals or find a common denominator, but cross-multiplication is usually efficient for pair-wise comparison.
The lower bound is $\frac{2}{7} \approx 0.2857$.
The upper bound is $\frac{7}{9} \approx 0.7778$.

Let’s test option B, $\frac{11}{17} \approx 0.6471$.
Is $\frac{11}{17} > \frac{2}{7}$? Compare $11 \times 7$ with $17 \times 2$. $77 > 34$. Yes, $\frac{11}{17} > \frac{2}{7}$.
Is $\frac{11}{17} < \frac{7}{9}$? Compare $11 \times 9$ with $17 \times 7$. $99 < 119$. Yes, $\frac{11}{17} < \frac{7}{9}$. Since $\frac{11}{17}$ satisfies both conditions, it is the correct answer. Let's quickly check other options: A) $\frac{15}{19} \approx 0.7895$. Is $\frac{15}{19} < \frac{7}{9}$? Compare $15 \times 9 = 135$ with $19 \times 7 = 133$. $135 > 133$, so $\frac{15}{19} > \frac{7}{9}$. Incorrect.
C) $\frac{11}{14} \approx 0.7857$. Is $\frac{11}{14} < \frac{7}{9}$? Compare $11 \times 9 = 99$ with $14 \times 7 = 98$. $99 > 98$, so $\frac{11}{14} > \frac{7}{9}$. Incorrect.
D) $\frac{17}{21} \approx 0.8095$. Is $\frac{17}{21} < \frac{7}{9}$? Compare $17 \times 9 = 153$ with $21 \times 7 = 147$. $153 > 147$, so $\frac{17}{21} > \frac{7}{9}$. Incorrect.

To compare two fractions $\frac{a}{b}$ and $\frac{c}{d}$ (with $b, d > 0$), compare $ad$ and $bc$. If $ad > bc$, then $\frac{a}{b} > \frac{c}{d}$. If $ad < bc$, then $\frac{a}{b} < \frac{c}{d}$.

Estimating decimal values of fractions can help quickly eliminate options, especially in objective tests, but precise comparison using cross-multiplication is more accurate.
$\frac{2}{7} \approx 0.28$
$\frac{7}{9} \approx 0.78$
A) $\frac{15}{19} \approx \frac{15}{20} = 0.75$. Closer check needed. (Actual: 0.789)
B) $\frac{11}{17} \approx \frac{11}{16} = 0.6875$ or $\frac{10}{16} = 0.625$. Looks promising. (Actual: 0.647)
C) $\frac{11}{14} \approx \frac{11}{15} \approx 0.73$. Closer check needed. (Actual: 0.786)
D) $\frac{17}{21} \approx \frac{17}{20} = 0.85$. Too high. (Actual: 0.810)

122. What is the decimal equivalent of the octal number 325.412?

What is the decimal equivalent of the octal number 325.412?

[amp_mcq option1=”5195.213″ option2=”339.5195″ option3=”5195.339″ option4=”213.5195″ correct=”option4″]

This question was previously asked in

UPSC CBI DSP LDCE – 2023

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To convert an octal number (base 8) to a decimal number (base 10), we multiply each digit by the corresponding power of 8 and sum the results.
For the integer part (325):
3 * 8^2 + 2 * 8^1 + 5 * 8^0
= 3 * 64 + 2 * 8 + 5 * 1
= 192 + 16 + 5
= 213

For the fractional part (.412):
4 * 8^-1 + 1 * 8^-2 + 2 * 8^-3
= 4/8 + 1/64 + 2/512
= 0.5 + 0.015625 + 0.00390625
= 0.51953125

Combining the parts: 213.51953125…
Looking at the options, 213.5195 is the closest representation of the decimal equivalent.

Octal to decimal conversion involves summing the products of each digit and the base (8) raised to the power of its position (positive for integer part, negative for fractional part).

The position of a digit to the left of the radix point starts at 0 and increases by 1 for each position further left. The position of a digit to the right of the radix point starts at -1 and decreases by 1 for each position further right.

123. A fair coin is tossed three times and the outcomes are noted. What is

A fair coin is tossed three times and the outcomes are noted. What is the probability of getting exactly two heads ?

[amp_mcq option1=”$\frac{2}{3}$” option2=”$\frac{1}{2}$” option3=”$\frac{5}{8}$” option4=”$\frac{3}{8}$” correct=”option4″]

This question was previously asked in

UPSC CAPF – 2024

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The correct option is D.

A fair coin is tossed three times.
Each toss has two possible outcomes: Heads (H) or Tails (T).
Since the tosses are independent, the total number of possible outcomes for three tosses is 2 * 2 * 2 = 2³ = 8.
The sample space (set of all possible outcomes) is:
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

We are looking for the probability of getting exactly two heads.
The outcomes with exactly two heads are:
HHT
HTH
THH
There are 3 favourable outcomes.

The probability of an event is defined as (Number of favourable outcomes) / (Total number of possible outcomes).
Probability of getting exactly two heads = (Number of outcomes with exactly two heads) / (Total number of outcomes)
Probability = 3 / 8.

For n independent coin tosses, the probability of getting exactly k heads can be calculated using the binomial probability formula: P(X=k) = C(n, k) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successes (heads), p is the probability of success on a single trial (probability of getting a head), and C(n, k) is the binomial coefficient (n choose k).
In this case, n=3 (three tosses), k=2 (exactly two heads), and p=0.5 (probability of getting a head with a fair coin).
P(X=2) = C(3, 2) * (0.5)² * (1-0.5)³⁻²
P(X=2) = C(3, 2) * (0.5)² * (0.5)¹
P(X=2) = C(3, 2) * (0.5)³
C(3, 2) = 3! / (2! * (3-2)!) = 3! / (2! * 1!) = (3 * 2 * 1) / ((2 * 1) * 1) = 3.
(0.5)³ = 0.5 * 0.5 * 0.5 = 0.125 = 1/8.
P(X=2) = 3 * (1/8) = 3/8.
This confirms the result obtained by listing the outcomes.

124. A group of five people consisting of a couple are to be seated on a ro

A group of five people consisting of a couple are to be seated on a round table for a meeting. What is the total number of ways in which the seating arrangement can be made so that the couple do NOT sit next to each other ?

[amp_mcq option1=”24″ option2=”18″ option3=”12″ option4=”6″ correct=”option3″]

This question was previously asked in

UPSC CAPF – 2024

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The correct option is C.

There are 5 people in total to be seated around a round table.
Total number of ways to arrange 5 distinct people around a round table is (5-1)! = 4! = 24.

We want to find the number of ways the couple do NOT sit next to each other. This can be found by subtracting the number of arrangements where the couple *do* sit together from the total number of arrangements.

To find the number of arrangements where the couple sit together, treat the couple as a single unit.
Now we are arranging 4 units around the table: the couple unit + the other 3 individuals.
The number of ways to arrange these 4 units around a round table is (4-1)! = 3! = 6.
Within the couple unit, the two individuals (let’s call them P1 and P2) can sit in two ways: P1-P2 or P2-P1. This can be done in 2! = 2 ways.

So, the total number of arrangements where the couple sit together is the number of ways to arrange the 4 units multiplied by the number of ways the couple can be arranged within their unit: 6 * 2 = 12.

The number of ways the couple do NOT sit next to each other is:
Total arrangements – Arrangements where the couple sit together
= 24 – 12 = 12.

For linear arrangements, the total ways for n people is n!. The ways for a specific couple to sit together is (n-1)! * 2!. The ways for them not to sit together is n! – (n-1)! * 2! = n! – 2(n-1)! = n(n-1)! – 2(n-1)! = (n-2)(n-1)!.
For round table arrangements, the total ways for n people is (n-1)!. The ways for a specific couple to sit together is (n-2)! * 2!. The ways for them not to sit together is (n-1)! – (n-2)! * 2! = (n-1)(n-2)! – 2(n-2)! = (n-1-2)(n-2)! = (n-3)(n-2)!.
In this case, n=5.
Total round table arrangements = (5-1)! = 4! = 24.
Couple sit together = (5-2)! * 2! = 3! * 2 = 6 * 2 = 12.
Couple do not sit together = (5-3)(5-2)! = 2 * 3! = 2 * 6 = 12.
The formula matches the calculation: (n-3)(n-2)! = (5-3)(5-2)! = 2 * 3! = 12.

125. Out of a class of 100 students, 25 play at least cricket and football,

Out of a class of 100 students, 25 play at least cricket and football, 15 play at least cricket and hockey, 12 play at least football and hockey and 10 play all the three sports. The number of students playing cricket, football and hockey are 50, 37 and 22, respectively. The number of students who do NOT play any of the three sports is

[amp_mcq option1=”33″ option2=”23″ option3=”27″ option4=”30″ correct=”option1″]

This question was previously asked in

UPSC CAPF – 2024

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The correct option is A.

Let C, F, and H represent the sets of students who play Cricket, Football, and Hockey, respectively.
Total students = 100.
Given:
|C| = 50
|F| = 37
|H| = 22
|C ∩ F| = 25 (students playing at least cricket and football implies the intersection)
|C ∩ H| = 15 (students playing at least cricket and hockey implies the intersection)
|F ∩ H| = 12 (students playing at least football and hockey implies the intersection)
|C ∩ F ∩ H| = 10 (students playing all three sports)

We need to find the number of students who do NOT play any of the three sports. This is given by Total students – |C ∪ F ∪ H|.
We use the Principle of Inclusion-Exclusion for three sets:
|C ∪ F ∪ H| = |C| + |F| + |H| – (|C ∩ F| + |C ∩ H| + |F ∩ H|) + |C ∩ F ∩ H|
Substitute the given values:
|C ∪ F ∪ H| = 50 + 37 + 22 – (25 + 15 + 12) + 10
|C ∪ F ∪ H| = 109 – (52) + 10
|C ∪ F ∪ H| = 109 – 52 + 10
|C ∪ F ∪ H| = 57 + 10 = 67.

The number of students who play at least one sport is 67.
The number of students who do NOT play any of the three sports = Total students – |C ∪ F ∪ H|
Number of students who do not play any sport = 100 – 67 = 33.

The phrasing “at least cricket and football” sometimes can be ambiguous. In standard set theory problems like this, it usually refers to the size of the intersection C ∩ F. If it meant only those who play exactly cricket and football (and not hockey), the numbers would be derived differently (e.g., |C ∩ F| – |C ∩ F ∩ H| would be the number playing *only* C and F). However, given the standard structure of such problems and the options, the interpretation used (intersection size) is almost certainly the intended one.
We can also visualize this with a Venn diagram. The value 10 goes in the center. The values for exactly two sports are:
Only C and F = |C ∩ F| – |C ∩ F ∩ H| = 25 – 10 = 15
Only C and H = |C ∩ H| – |C ∩ F ∩ H| = 15 – 10 = 5
Only F and H = |F ∩ H| – |C ∩ F ∩ H| = 12 – 10 = 2
The values for exactly one sport are:
Only C = |C| – (Only C&F) – (Only C&H) – (C&F&H) = 50 – 15 – 5 – 10 = 20
Only F = |F| – (Only C&F) – (Only F&H) – (C&F&H) = 37 – 15 – 2 – 10 = 10
Only H = |H| – (Only C&H) – (Only F&H) – (C&F&H) = 22 – 5 – 2 – 10 = 5
Total playing at least one sport = (Only C) + (Only F) + (Only H) + (Only C&F) + (Only C&H) + (Only F&H) + (C&F&H)
= 20 + 10 + 5 + 15 + 5 + 2 + 10 = 67.
Number not playing any sport = 100 – 67 = 33. This confirms the result.

126. A shopkeeper sells two items, A and B. Item B’s cost price is twice as

A shopkeeper sells two items, A and B. Item B’s cost price is twice as that of item A. The shopkeeper sells item A at 10% profit and item B at 20% profit. Which one of the following is the value of net profit ?

[amp_mcq option1=”15%” option2=”13.33%” option3=”18%” option4=”16.66%” correct=”option4″]

This question was previously asked in

UPSC CAPF – 2024

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The correct option is D.

Let the cost price of item A be CP_A.
Let the cost price of item B be CP_B.
Given: CP_B = 2 * CP_A.
For calculation, let CP_A = ₹100.
Then CP_B = 2 * ₹100 = ₹200.
Total Cost Price = CP_A + CP_B = ₹100 + ₹200 = ₹300.
Item A is sold at 10% profit.
Profit on A = 10% of CP_A = 10% of ₹100 = ₹10.
Selling Price of A (SP_A) = CP_A + Profit on A = ₹100 + ₹10 = ₹110.
Item B is sold at 20% profit.
Profit on B = 20% of CP_B = 20% of ₹200 = ₹40.
Selling Price of B (SP_B) = CP_B + Profit on B = ₹200 + ₹40 = ₹240.
Total Selling Price = SP_A + SP_B = ₹110 + ₹240 = ₹350.
Net Profit = Total Selling Price – Total Cost Price = ₹350 – ₹300 = ₹50.
Net Profit Percentage = (Net Profit / Total Cost Price) * 100
Net Profit Percentage = (₹50 / ₹300) * 100
Net Profit Percentage = (1/6) * 100 = 100/6 = 50/3 = 16.66…%.

Alternatively, using variables:
Let CP_A = x. Then CP_B = 2x. Total CP = 3x.
SP_A = x + 0.10x = 1.1x.
SP_B = 2x + 0.20(2x) = 2x + 0.4x = 2.4x.
Total SP = 1.1x + 2.4x = 3.5x.
Net Profit = Total SP – Total CP = 3.5x – 3x = 0.5x.
Net Profit Percentage = (0.5x / 3x) * 100 = (0.5 / 3) * 100 = (1/6) * 100 = 16.66…%.
This is a weighted average profit. Item B has twice the cost price of A, so its profit contributes twice as much to the total profit in absolute terms. The overall profit percentage is closer to the profit percentage of the more expensive item.

127. The sum of the ages of A and B (in years) is 22. The product of their

The sum of the ages of A and B (in years) is 22. The product of their ages two years back was 77. Which one of the following is the value of the difference of their current ages ?

[amp_mcq option1=”2″ option2=”3″ option3=”4″ option4=”5″ correct=”option3″]

This question was previously asked in

UPSC CAPF – 2024

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The correct option is C.

Let the current ages of A and B be ‘a’ and ‘b’ years, respectively.
Given: a + b = 22.
Two years back, their ages were (a-2) and (b-2).
Given: (a-2)(b-2) = 77.
Expanding the second equation: ab – 2a – 2b + 4 = 77
ab – 2(a + b) + 4 = 77.
Substitute the first equation (a + b = 22) into this:
ab – 2(22) + 4 = 77
ab – 44 + 4 = 77
ab – 40 = 77
ab = 117.
We need to find the difference of their current ages, which is |a – b|.
We use the identity: (a – b)² = (a + b)² – 4ab.
Substituting the values we have:
(a – b)² = (22)² – 4(117)
(a – b)² = 484 – 468
(a – b)² = 16.
Therefore, |a – b| = √16 = 4.
The difference in their current ages is 4 years.

We can also find the individual ages. We have a + b = 22 and ab = 117. This means ‘a’ and ‘b’ are the roots of the quadratic equation x² – (a+b)x + ab = 0, which is x² – 22x + 117 = 0.
Factoring the quadratic: (x – 9)(x – 13) = 0.
The roots are x = 9 and x = 13.
So, the ages are 9 and 13 years.
Let’s check the conditions:
Sum of ages = 9 + 13 = 22 (Correct).
Ages two years back were 9-2 = 7 and 13-2 = 11.
Product of ages two years back = 7 * 11 = 77 (Correct).
The difference in their current ages is |13 – 9| = 4.

128. In a partnership firm, A invests $\frac{1}{6}$th of the total invest

In a partnership firm, A invests $\frac{1}{6}$th of the total investment for $\frac{1}{6}$th of the tenure. B invests $\frac{1}{3}$rd of the total investment for $\frac{1}{3}$rd of the tenure while C invests the remaining part for the full duration. Out of total profit of ₹46,00,000, what shall be C’s share ?

[amp_mcq option1=”₹30,00,000″ option2=”₹32,00,000″ option3=”₹34,00,000″ option4=”₹36,00,000″ correct=”option4″]

This question was previously asked in

UPSC CAPF – 2024

Download PDF Attempt Online

C’s share out of the total profit shall be ₹36,00,000.

In a partnership, the profit share of each partner is proportional to the product of their investment and the duration for which the investment was made.
Let the total investment be $I$ and the total tenure be $T$.
A’s investment $I_A = \frac{1}{6} I$. A’s tenure $T_A = \frac{1}{6} T$.
A’s profit share is proportional to $I_A \times T_A = (\frac{1}{6} I) \times (\frac{1}{6} T) = \frac{1}{36} IT$.

B’s investment $I_B = \frac{1}{3} I$. B’s tenure $T_B = \frac{1}{3} T$.
B’s profit share is proportional to $I_B \times T_B = (\frac{1}{3} I) \times (\frac{1}{3} T) = \frac{1}{9} IT$.

C invests the remaining part of the investment for the full duration.
C’s investment $I_C = \text{Total Investment} – I_A – I_B$.
$I_C = I – \frac{1}{6} I – \frac{1}{3} I = I – (\frac{1}{6} + \frac{2}{6}) I = I – \frac{3}{6} I = I – \frac{1}{2} I = \frac{1}{2} I$.
C’s tenure $T_C = T$ (full duration).
C’s profit share is proportional to $I_C \times T_C = (\frac{1}{2} I) \times T = \frac{1}{2} IT$.

The ratio of the profit shares of A, B, and C is the ratio of their (Investment $\times$ Time) products:
Ratio = $\frac{1}{36} IT : \frac{1}{9} IT : \frac{1}{2} IT$.
Dividing by $IT$ (assuming $I, T > 0$):
Ratio = $\frac{1}{36} : \frac{1}{9} : \frac{1}{2}$.
To simplify this ratio, multiply by the Least Common Multiple (LCM) of the denominators (36, 9, 2), which is 36.
Ratio = $36 \times \frac{1}{36} : 36 \times \frac{1}{9} : 36 \times \frac{1}{2}$
Ratio = $1 : 4 : 18$.

The sum of the ratio parts is $1 + 4 + 18 = 23$.
The total profit is ₹46,00,000.
C’s share of the profit is the total profit multiplied by C’s ratio part divided by the sum of ratio parts.
C’s share = $\frac{18}{23} \times ₹46,00,000$.
$\frac{46,00,000}{23} = 2,00,000$.
C’s share = $18 \times ₹2,00,000 = ₹36,00,000$.

The fundamental principle of profit sharing in a partnership is that profits are distributed in proportion to the capital invested and the time period for which the capital was invested. If investments are $I_1, I_2, …, I_n$ for durations $T_1, T_2, …, T_n$, the profit ratio is $I_1T_1 : I_2T_2 : … : I_nT_n$. The sum of the investments must equal the total investment, and the durations considered for calculating the ratio should be consistent (e.g., all in months or all in years, or simply relative units if the total tenure is used as a base like here).

129. A train travelling at a speed of 60 km/hr crosses a platform in 20 sec

A train travelling at a speed of 60 km/hr crosses a platform in 20 seconds. The same train crosses a person who is walking at a speed of 6 km/hr in the same direction as that of the train in 12 seconds. What is the length of the train and that of the platform, respectively ?

[amp_mcq option1=”160 m and 153.33 m” option2=”170 m and 166.66 m” option3=”180 m and 153.33 m” option4=”180 m and 170 m” correct=”option3″]

This question was previously asked in

UPSC CAPF – 2024

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The length of the train is 180 m and that of the platform is approximately 153.33 m.

Let the length of the train be $L_T$ (in meters) and the length of the platform be $L_P$ (in meters).
The speed of the train is given as 60 km/hr. Let’s convert this to meters per second (m/s).
Speed in m/s = Speed in km/hr $\times \frac{5}{18}$.
$V_T = 60 \times \frac{5}{18} = \frac{300}{18} = \frac{50}{3}$ m/s.

Case 1: Train crosses a platform in 20 seconds.
When a train crosses a platform, the total distance covered by the train is the sum of its own length and the length of the platform ($L_T + L_P$).
Distance = Speed $\times$ Time.
$L_T + L_P = V_T \times 20$.
$L_T + L_P = \frac{50}{3} \times 20 = \frac{1000}{3}$ meters. (Equation 1)

Case 2: Train crosses a person walking at 6 km/hr in the same direction in 12 seconds.
The speed of the person is 6 km/hr. Convert this to m/s.
$V_P = 6 \times \frac{5}{18} = \frac{30}{18} = \frac{5}{3}$ m/s.
When the train crosses a person moving in the same direction, we use the relative speed.
Relative speed = Speed of train – Speed of person (since they are in the same direction).
Relative speed $= V_T – V_P = \frac{50}{3} – \frac{5}{3} = \frac{45}{3} = 15$ m/s.
The distance covered by the train relative to the person is the length of the train ($L_T$).
Distance = Relative Speed $\times$ Time.
$L_T = 15 \times 12 = 180$ meters.

Now that we have the length of the train ($L_T = 180$ m), we can use Equation 1 to find the length of the platform ($L_P$).
$L_T + L_P = \frac{1000}{3}$
$180 + L_P = \frac{1000}{3}$
$L_P = \frac{1000}{3} – 180 = \frac{1000 – 180 \times 3}{3} = \frac{1000 – 540}{3} = \frac{460}{3}$ meters.
Convert $\frac{460}{3}$ to a decimal: $460 \div 3 \approx 153.33$.
So, $L_T = 180$ m and $L_P \approx 153.33$ m.
The question asks for the length of the train and that of the platform, respectively.

Key concepts used: conversion of speed units (km/hr to m/s), calculating distance when a train crosses an object (stationary or moving), and relative speed for objects moving in the same or opposite directions. When a train crosses a stationary object of negligible length (like a person or a pole), the distance covered is the length of the train. When it crosses an object of considerable length (like a platform or another train), the distance covered is the sum of the lengths. Relative speed is the difference of speeds for objects moving in the same direction and the sum of speeds for objects moving in opposite directions.

130. 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

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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).
$P(6, 2) = \frac{6!}{(6-2)!} = \frac{6!}{4!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} = 6 \times 5 = 30$.