11.3 2D/3D Mensuration & Banking Permutation, Combination, Probability

11.3 2D/3D Mensuration & Banking Permutation, Combination, Probability

Mensuration, Combinatorics, and Probability form the applied mathematics foundation of the SBI Clerk Examination. Mensuration principles govern tangible asset valuation: bank loan officers survey land parcels offered as mortgage collateral, estimate building plinth areas for home loan disbursements, evaluate branch remodeling floor surfaces, and calculate cubic strongroom vault capacities for bullion storage. Concurrently, Permutation & Combination (P&C) and Probability govern banking security and risk analytics: designing alphanumeric authentication credentials, ATM PIN security parameters, teller rotation committee formations, and modeling probability distributions in credit portfolio defaults.

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1. 2D Mensuration: Formulas, Pathways & Composite Figures

Two-dimensional geometry measures boundary perimeters and bounded surface areas across planar geometric figures.

Rectangles, Squares, and Pathway Geometry

1. Rectangle of length $l$ and breadth $b$: $\text{Area} = l \times b, \qquad \text{Perimeter} = 2(l + b), \qquad \text{Diagonal } (d) = \sqrt{l^2 + b^2}$

2. Square of side $a$: $\text{Area} = a^2 = \frac{1}{2}d^2, \qquad \text{Perimeter} = 4a, \qquad \text{Diagonal } (d) = a\sqrt{2}$

3. Pathway Problems on Rectangular Fields:

   • Uniform Outer Path of width $w$: $\text{Area of Path} = (l + 2w)(b + 2w) - lb = 2w(l + b + 2w)$
   • Uniform Inner Path of width $w$: $\text{Area of Path} = lb - (l - 2w)(b - 2w) = 2w(l + b - 2w)$
   • Central Crossing Paths of width $w$ running parallel to sides: $\text{Area of Crossroads} = (l \times w) + (b \times w) - w^2 = w(l + b - w)$

Triangles and Circles Reference

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2. 3D Mensuration: Solid Geometry & Recasting Invariance

Three-dimensional mensuration evaluates the physical space enclosed by solids (volume) and the exterior surface coverage (lateral/curved and total surface area).

3D Solids Mathematical Formulations

• Cube (Edge $a$): $\text{Volume} = a^3, \quad \text{Lateral Surface Area (LSA)} = 4a^2, \quad \text{Total Surface Area (TSA)} = 6a^2, \quad \text{Diagonal} = a\sqrt{3}$

• Cuboid (Dimensions $l, b, h$): $\text{Volume} = lbh, \quad \text{LSA (Area of 4 walls)} = 2h(l + b), \quad \text{TSA} = 2(lb + bh + hl), \quad \text{Diagonal} = \sqrt{l^2 + b^2 + h^2}$

• Right Circular Cylinder (Radius $r$, Height $h$): $\text{Volume} = \pi r^2 h, \quad \text{Curved Surface Area (CSA)} = 2\pi rh, \quad \text{TSA} = 2\pi r(r + h)$

• Right Circular Cone (Radius $r$, Height $h$, Slant Height $l = \sqrt{r^2 + h^2}$): $\text{Volume} = \frac{1}{3}\pi r^2 h, \quad \text{CSA} = \pi rl, \quad \text{TSA} = \pi r(r + l)$

• Solid Sphere (Radius $r$): $\text{Volume} = \frac{4}{3}\pi r^3, \quad \text{Surface Area} = 4\pi r^2$

• Solid Hemisphere (Radius $r$): $\text{Volume} = \frac{2}{3}\pi r^3, \quad \text{CSA} = 2\pi r^2, \quad \text{TSA} = 3\pi r^2$

The Recasting and Melting Invariance Principle

When a physical metal block (such as an ingot from a bank vault) is melted and recast into smaller units, the total volume remains completely invariant, even though total surface area changes substantially:

$\text{Volume of Original Solid} = \sum \text{Volumes of Recast Solids}$

$n \times \text{Volume of One Small Recast Unit} = \text{Volume of Original Solid}$

$n = \frac{\text{Volume of Original Solid}}{\text{Volume of One Recast Unit}}$

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3. Permutations vs Combinations: The Decision Framework

Combinatorics counts configurations without listing them individually. The fundamental decision when analyzing a problem is determining whether sequence or order matters.

Counting Problem: Select r items from n distinct items
      |
Does ORDER matter?
     /  \
   YES   NO
   /       \
PERMUTATION   COMBINATION
  nPr           nCr
(Arrangements,  (Selections, Groups,
 Codes, Ranks)   Committees, Cards)

Fundamental Counting Axioms

1. Multiplication Rule (AND): If an operation can be performed in $m$ ways, and following this a second operation can be performed in $n$ ways, the sequential occurrence of both operations can be executed in $m \times n$ distinct ways.
2. Addition Rule (OR): If mutually exclusive alternatives exist such that an event can happen via process 1 in $m$ ways OR via process 2 in $n$ ways, the total number of ways is $m + n$.

Mathematical Formulas

• Permutation (Ordered Arrangement): $^nP_r = \frac{n!}{(n - r)!}$

• Combination (Unordered Selection): $^nC_r = \frac{n!}{r!(n - r)!}$

• Key Combination Properties: $^nC_r = ^nC_{n - r}, \qquad ^nC_0 = ^nC_n = 1, \qquad ^nC_1 = n, \qquad ^nC_r + ^nC_{r - 1} = ^{n+1}C_r$

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4. Word Arrangement Techniques & The Tie/String Method

A standard question archetype in the SBI Clerk exam asks for the number of distinct arrangements of letters in an English word under vowel/consonant constraints.

Formula for Permutations with Repeated Elements

If a word contains $n$ total letters where letter 1 appears $p$ times, letter 2 appears $q$ times, and letter 3 appears $r$ times: $\text{Total Distinct Arrangements} = \frac{n!}{p! \times q! \times r!}$

The Tie/String Method (Vowels Always Together)

1. Bundle all the required vowels into a single composite entity: {V_1, V_2, ...}.
2. Count the total units: (remaining consonants) $+ 1$ (the composite vowel unit).
3. Compute the external arrangements of these composite units, dividing by factorials of repeating consonants.
4. Multiply by internal permutations of the vowels within the tie bundle, dividing by factorials of repeating vowels.

$\text{Total Ways} = (\text{Arrangements of Units}) \times (\text{Internal Permutations of Vowels})$

The 'Never Together' Subtraction Shortcut

To determine arrangements where specific letters never appear adjacent to one another: $\text{Arrangements (Never Together)} = \text{Total Unrestricted Arrangements} - \text{Arrangements (Always Together)}$

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5. Classical Probability Models in Banking Exams

Probability measures the likelihood of a specific event occurring relative to the exhaustive sample space of all possible outcomes:

$\text{Probability } P(E) = \frac{n(E)}{n(S)} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Exhaustive Outcomes}}, \qquad 0 \le P(E) \le 1$

The Complementary Probability Rule

In problems featuring conditions such as "at least one", calculating favorable combinations directly requires summing multiple cases. Instead, solve via the complement:

$P(\text{At least one } E) = 1 - P(\text{No } E)$

Standard Banking Probability Arenas

1. Standard 52-Card Deck:

   • 4 Suits (13 cards each): Spades (♠, black), Clubs (♣, black), Hearts (♥, red), Diamonds (♦, red).
   • 12 Face / Court Cards: 4 Jacks, 4 Queens, 4 Kings (3 per suit).
   • 16 Honor Cards: 4 Aces, 4 Kings, 4 Queens, 4 Jacks (4 per suit).

2. Urn & Token Selection (Without Replacement): When $r$ tokens are selected simultaneously from an urn containing $N$ total tokens ($N_1$ red, $N_2$ green, $N_3$ blue): $n(S) = ^NC_r$ $P(\text{all red}) = \frac{^{N_1}C_r}{^NC_r}, \qquad P(\text{no red}) = \frac{^{N - N_1}C_r}{^NC_r}$

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6. Step-by-Step Worked Problems

Problem 1: Metal Recasting Problem

Problem: A solid metallic cylinder of base radius 6 cm and height 16 cm is melted down and recast into identical spherical balls of diameter 4 cm. How many balls are formed?

1. Compute Cylinder Volume:
   $V_{\text{cyl}} = \pi R^2 H = \pi \times 6^2 \times 16 = \pi \times 36 \times 16 = 576\pi \text{ cm}^3$
2. Determine Sphere Parameters:
   Diameter $= 4 \text{ cm} \implies \text{radius } r = 2 \text{ cm}$.
3. Compute Volume of One Sphere:
   $V_{\text{sph}} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (2)^3 = \frac{32}{3}\pi \text{ cm}^3$
4. Apply Volume Invariance:
   $n = \frac{V_{\text{cyl}}}{V_{\text{sph}}} = \frac{576\pi}{\frac{32}{3}\pi} = \frac{576 \times 3}{32} = 18 \times 3 = \mathbf{54 \text{ balls}}$

Problem 2: Urn Selection with 'At Least One'

Problem: A cash collection urn holds 4 red tokens, 5 green tokens, and 3 blue tokens (12 tokens total). If 2 tokens are drawn at random without replacement, find the probability that at least one green token is selected.

1. Determine Total Sample Space:
   $n(S) = ^{12}C_2 = \frac{12 \times 11}{2 \times 1} = 66$
2. Determine Non-Green Tokens:
   $\text{Non-Green Tokens} = 4 \text{ red} + 3 \text{ blue} = 7 \text{ tokens}$
3. Calculate Outcomes with Zero Green Tokens:
   $n(\text{no green}) = ^7C_2 = \frac{7 \times 6}{2 \times 1} = 21$
4. Compute Complementary Probability:
   $P(\text{no green}) = \frac{21}{66} = \frac{7}{22}$
5. Subtract from Unity:
   $P(\text{at least one green}) = 1 - P(\text{no green}) = 1 - \frac{7}{22} = \mathbf{\frac{15}{22}}$

Problem 3: Committee Selection With a Restriction

Problem: From 6 men and 4 women, a committee of 5 is to be formed with at least 2 women. In how many ways can this be done?

1. Split into cases by the number of women:
   • 2 women and 3 men: $^4C_2 \times {}^6C_3 = 6 \times 20 = 120$
   • 3 women and 2 men: $^4C_3 \times {}^6C_2 = 4 \times 15 = 60$
   • 4 women and 1 man: $^4C_4 \times {}^6C_1 = 1 \times 6 = 6$
2. Add the mutually exclusive cases: $120 + 60 + 6 = \mathbf{186}$ committees.
3. Check with the complement: All committees number $^{10}C_5 = 252$. Subtract those with no woman ($^6C_5 = 6$) and exactly one woman ($^4C_1 \times {}^6C_4 = 4 \times 15 = 60$): $252 - 66 = 186$.