Connecting the Dots

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CLASS VII Mathematics ~6 marks/year Ch 13 of 15
Connecting the Dots

Class 7 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • This chapter is an introduction to permutations and combinations — the maths of counting arrangements without listing every case.
  • The big idea is the Fundamental Principle of Counting: if one task can happen in $m$ ways and a second in $n$ ways, both together happen in $m\times n$ ways.
  • You learn to count routes on a grid, handshakes, diagonals of a polygon, and line segments formed by joining dots.
  • The handshake / line-segment number for $n$ dots is $\dfrac{n(n-1)}{2}$ — a formula that appears again and again.
  • Counting choices where order matters (arrangements) is different from where order doesn't matter (selections).
  • You meet tree diagrams and systematic listing as safe ways to count small cases and to check formulas.
  • Weightage: ~6 marks/year — usually a counting-routes or handshake/diagonal problem plus 1–2 short objective questions.
Detailed Notes

1. What "connecting the dots" really means

Imagine some dots drawn on paper. You can join them with lines, walk along them as a path, or pick a few of them as a team. Each of these is a counting question: how many? How many lines can I draw? How many different paths reach the corner? How many teams can I form? This chapter teaches you to answer such questions cleverly, without drawing or listing everything.

The whole subject is called combinatorics — the art of counting. Its first and most useful rule is the multiplication principle, and almost every problem in this chapter is built from it.

2. The Fundamental Principle of Counting

Suppose you are doing two things one after another. If the first thing can be done in $m$ ways, and for each of those the second can be done in $n$ ways, then the two together can be done in $m\times n$ ways.

If a job has two stages with $m$ and $n$ choices, total ways $=m\times n$. For three stages with $m,n,p$ choices, total $=m\times n\times p$, and so on.

This works because each choice at stage one "branches" into all the choices at stage two. Multiplying the branches counts every combination exactly once.

Worked example — dressing up

Riya has $3$ shirts and $4$ skirts. How many shirt-and-skirt outfits can she wear?

Stage 1: pick a shirt — $3$ ways. Stage 2: pick a skirt — $4$ ways. By the principle, total outfits $=3\times4=12$.

Worked example — number plates

A code has one letter (A, B or C) followed by one digit (from $0$ to $9$). How many codes are possible?

Letters: $3$ ways. Digits: $10$ ways ($0,1,\dots,9$). Total $=3\times10=30$ codes.

3. Tree diagrams — seeing the branches

A tree diagram draws every choice as a branch. It is slow but trustworthy, and it makes the multiplication principle visible.

Worked example — coin tossed twice

Toss a coin twice. First toss branches into Heads (H) and Tails (T). From each, the second toss again branches into H and T. The leaves are: HH, HT, TH, TT — that is $2\times2=4$ outcomes, matching the principle.

Use a tree (or a neat list) whenever the numbers are small or you are unsure of a formula. If your formula and your list disagree, the list is usually right.

4. Order matters vs order does not matter

This is the single most important distinction in the chapter.

  • Order matters (arrangement): "first, second, third" are different roles. Captain then vice-captain is not the same as vice-captain then captain.
  • Order does not matter (selection): just a group, no roles. The team $\{$Asha, Bina$\}$ is the same as $\{$Bina, Asha$\}$.
Worked example — feel the difference

From $3$ friends A, B, C choose $2$.

As a team (order ignored): AB, AC, BC — that is $3$ selections.

As captain & vice-captain (order counts): AB, BA, AC, CA, BC, CB — that is $6$ arrangements.

Notice arrangements $=2\times$ selections here, because each pair can be ordered in $2$ ways.

5. Joining dots — line segments

Take $n$ dots so that no three lie on a straight line. A line segment needs two dots, and the order does not matter (the segment from P to Q is the same as Q to P).

Count cleverly: the first endpoint can be any of $n$ dots, the second any of the remaining $n-1$. That gives $n(n-1)$ ordered pairs — but each segment is counted twice (PQ and QP). So divide by $2$.

Number of line segments joining $n$ dots (no three in a line) $=\dfrac{n(n-1)}{2}$.
Worked example — segments from 5 dots

With $n=5$: segments $=\dfrac{5\times4}{2}=\dfrac{20}{2}=10$. Check by listing pairs from A,B,C,D,E: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE — exactly $10$.

6. The handshake problem

If $n$ people each shake hands once with every other person, the number of handshakes is exactly the same as the number of line segments — because each handshake is a "pair of people".

Handshakes among $n$ people $=\dfrac{n(n-1)}{2}$.
Worked example — a party of 8

$8$ guests, everyone shakes hands once. Total $=\dfrac{8\times7}{2}=\dfrac{56}{2}=28$ handshakes.

Worked example — work backwards

A group had $15$ handshakes. How many people? Solve $\dfrac{n(n-1)}{2}=15\Rightarrow n(n-1)=30$. Since $6\times5=30$, we get $n=6$ people.

7. Diagonals of a polygon

A polygon with $n$ vertices: the segments joining pairs of vertices number $\dfrac{n(n-1)}{2}$. But $n$ of those are the sides, not diagonals. So subtract $n$.

Number of diagonals of an $n$-sided polygon $=\dfrac{n(n-1)}{2}-n=\dfrac{n(n-3)}{2}$.
Worked example — pentagon and hexagon

Pentagon $n=5$: diagonals $=\dfrac{5(5-3)}{2}=\dfrac{5\times2}{2}=5$.

Hexagon $n=6$: diagonals $=\dfrac{6(6-3)}{2}=\dfrac{6\times3}{2}=9$.

A triangle ($n=3$) gives $\dfrac{3\times0}{2}=0$ diagonals — correct, a triangle has none.

8. Counting paths on a grid

Picture a grid of streets. You start at the bottom-left dot and must reach the top-right dot, moving only right or up. How many shortest paths are there?

The neat trick: write at each dot the number of ways to reach it. The start has $1$ way. Any other dot's number equals the sum of the number to its left and the number below it (since you arrive only from those two directions).

Worked example — a 2×2 block of streets

Label corners on a $2\times2$ grid (3 dots along each side). Bottom row reads $1,1,1$ (only one way along an edge). Moving up, each interior dot adds left+below: the middle row becomes $1,2,3$ and the top row $1,3,6$. So the far corner has $\boldsymbol{6}$ shortest paths.

This "add the two behind you" idea is the same pattern that builds Pascal's triangle.

9. Pascal's triangle — the pattern of choices

Start with a $1$ at the top. Each new number is the sum of the two numbers above it (treat missing neighbours as $0$). The rows are:

$1$
$1\quad 1$
$1\quad 2\quad 1$
$1\quad 3\quad 3\quad 1$
$1\quad 4\quad 6\quad 4\quad 1$

These numbers tell you, for example, how many ways to choose objects from a set, and they match the grid-path counts. Patterns hidden inside include: the outer edges are all $1$; each row is symmetric (a mirror image); and each row adds up to a power of $2$ ($1,2,4,8,16,\dots$).

10. Counting arrangements in a row

How many ways can $n$ different objects be arranged in a line? Fill the first place ($n$ choices), then the second ($n-1$ left), then the third ($n-2$), and so on down to $1$.

Arrangements of $n$ different objects in a row $=n\times(n-1)\times(n-2)\times\cdots\times2\times1$ (called "$n$ factorial").
Worked example — 3 books on a shelf

$3$ books: $3\times2\times1=6$ arrangements. Listing for books P, Q, R: PQR, PRQ, QPR, QRP, RPQ, RQP — exactly $6$.

Worked example — 4 students in a queue

$4$ students: $4\times3\times2\times1=24$ different queues.

11. Putting it together — mixed counting

Worked example — meals

A thali offers $2$ breads, $3$ sabzis and $2$ sweets. Choosing one of each: $2\times3\times2=12$ different thalis.

Worked example — choosing then arranging

From $4$ runners, in how many ways can the gold and silver medals be given? Gold: $4$ ways; silver: $3$ remaining $=4\times3=12$. (Order matters, so this is an arrangement, not a selection.)

12. Common mistakes to avoid

  • Forgetting to divide by 2 for line segments/handshakes — PQ and QP are the same.
  • Mixing up arrangement and selection — ask "do roles/positions matter?" If yes, order counts.
  • Adding when you should multiply — use $+$ only for "either this OR that", and $\times$ for "this AND then that".
  • Counting sides as diagonals — remember to subtract the $n$ sides.
  • Listing carelessly — be systematic (alphabetical or numerical order) so nothing is missed or repeated.

13. Quick revision checklist

  • Multiplication principle: stages with $m,n,p\dots$ choices → multiply.
  • Line segments / handshakes for $n$ dots/people $=\dfrac{n(n-1)}{2}$.
  • Diagonals of an $n$-gon $=\dfrac{n(n-3)}{2}$.
  • Grid paths: each dot = (left) + (below); the corner gives the answer.
  • Arranging $n$ objects in a row $=n\times(n-1)\times\cdots\times1$.
  • "AND" → multiply; "OR" → add. Order matters → arrangement; order ignored → selection.
Practice MCQs
1. A canteen has $4$ types of bread and $3$ types of jam. The number of bread-and-jam combinations is:
  1. $7$
  2. $12$
  3. $1$
  4. $43$
Answer: (B) By the multiplication principle, $4\times3=12$.
2. The number of line segments that can be drawn joining $6$ dots (no three in a line) is:
  1. $6$
  2. $12$
  3. $15$
  4. $30$
Answer: (C) $\dfrac{6\times5}{2}=15$.
3. In a room, every two people shake hands once. If there are $28$ handshakes, the number of people is:
  1. $7$
  2. $8$
  3. $14$
  4. $28$
Answer: (B) $\dfrac{n(n-1)}{2}=28\Rightarrow n(n-1)=56=8\times7$, so $n=8$.
4. The number of diagonals of a hexagon (6 sides) is:
  1. $6$
  2. $9$
  3. $12$
  4. $15$
Answer: (B) $\dfrac{6(6-3)}{2}=\dfrac{18}{2}=9$.
5. How many different $2$-letter "words" (letters may repeat) can be made from A, B, C, D?
  1. $8$
  2. $12$
  3. $16$
  4. $4$
Answer: (C) Each place has $4$ choices: $4\times4=16$.
6. The number of ways to arrange $4$ different books in a row is:
  1. $4$
  2. $12$
  3. $16$
  4. $24$
Answer: (D) $4\times3\times2\times1=24$.
7. From $5$ players, gold and silver medals are awarded. The number of ways is:
  1. $10$
  2. $20$
  3. $25$
  4. $5$
Answer: (B) Order matters: $5\times4=20$.
8. The number of triangles that can be formed by joining $5$ dots (no three in a line) — i.e. choosing $3$ of them — is:
  1. $5$
  2. $10$
  3. $15$
  4. $20$
Answer: (B) Choosing $3$ from $5$ ignoring order gives $10$ (ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE).
9. To form a team (order not important) of $2$ from $4$ people, the number of teams is:
  1. $12$
  2. $8$
  3. $6$
  4. $4$
Answer: (C) AB, AC, AD, BC, BD, CD $=6$ teams.
10. A coin is tossed $3$ times. The number of possible outcomes is:
  1. $3$
  2. $6$
  3. $8$
  4. $9$
Answer: (C) $2\times2\times2=8$.
11. On a grid, you move only right or up from the bottom-left to the top-right corner of a $2\times2$ block. The number of shortest paths is:
  1. $2$
  2. $4$
  3. $6$
  4. $8$
Answer: (C) Adding left+below across the grid gives $6$ at the far corner.
12. The sum of the numbers in the row $1, 4, 6, 4, 1$ of Pascal's triangle is:
  1. $8$
  2. $15$
  3. $16$
  4. $32$
Answer: (C) $1+4+6+4+1=16=2^4$.
13. A number plate has $2$ letters from A, B, C followed by $1$ digit ($0$–$9$); letters may repeat. The total count is:
  1. $60$
  2. $90$
  3. $30$
  4. $13$
Answer: (B) $3\times3\times10=90$.
14. The number of sides plus the number of diagonals of a pentagon equals the number of segments joining its $5$ vertices, which is:
  1. $5$
  2. $8$
  3. $10$
  4. $15$
Answer: (C) $\dfrac{5\times4}{2}=10$ ($5$ sides $+5$ diagonals).
15. Which operation joins choices when you do "this AND then that"?
  1. Addition
  2. Subtraction
  3. Multiplication
  4. Division
Answer: (C) "AND then" stages are multiplied; "OR" choices are added.
Important Questions
Q1. State the Fundamental Principle of Counting and use it to find how many $3$-digit codes can be made from digits $1$–$5$ if repetition is allowed. (3 marks)
Answer: If one task can be done in $m$ ways and a second in $n$ ways, both can be done in $m\times n$ ways (extend by multiplying for more stages). Here each of the $3$ places has $5$ choices, so total codes $=5\times5\times5=125$.
Q2. Derive the formula for the number of line segments joining $n$ dots (no three collinear) and use it for $n=10$. (3 marks)
Answer: First endpoint $n$ ways, second $n-1$ ways, giving $n(n-1)$ ordered pairs; each segment counted twice, so divide by $2$: $\dfrac{n(n-1)}{2}$. For $n=10$: $\dfrac{10\times9}{2}=45$ segments.
Q3. Find the number of diagonals of an octagon ($8$ sides) and explain why we subtract the sides. (2 marks)
Answer: Segments joining $8$ vertices $=\dfrac{8\times7}{2}=28$. Of these, $8$ are sides, not diagonals, so diagonals $=28-8=20$ (or directly $\dfrac{8(8-3)}{2}=20$).
Q4. Explain the difference between an arrangement and a selection using $3$ friends A, B, C choosing $2$. (3 marks)
Answer: Selection (order ignored): AB, AC, BC $=3$. Arrangement (order matters, e.g. captain & vice-captain): AB, BA, AC, CA, BC, CB $=6$. Each selected pair can be ordered in $2$ ways, so arrangements $=2\times$ selections.
Q5. On a $3\times3$ grid of streets, count the shortest right/up paths from the bottom-left to the top-right corner. (3 marks)
Answer: Write $1$ along the bottom and left edges, then each interior dot = left + below. The rows become $1,1,1,1$ / $1,2,3,4$ / $1,3,6,10$ / $1,4,10,20$. The far corner shows $\boldsymbol{20}$ shortest paths.
Q6. How many ways can $5$ students stand in a row for a photo? (2 marks)
Answer: Fill positions one by one: $5\times4\times3\times2\times1=120$ ways.
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Large Numbers Around Us · Arithmetic Expressions · A Peek Beyond the Point · Expressions using Letter-Numbers · Parallel and Intersecting Lines · Number Play · A Tale of Three Intersecting Lines · Working with Fractions
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