Binomial Expansion of (a + bx)ⁿ for Positive Integer n

What is the coefficient of $x^4$ in the expansion of $(2 + 5x)^9$?

• $2\,520\,000$

  Correct. $\binom{9}{4} = 126$, $2^{9-4} = 2^5 = 32$, $5^4 = 625$, and $126 \times 32 \times 625 = 2\,520\,000$.

• B$6\,300\,000$

  This swaps which exponent belongs to which factor — computing $2^4 \times 5^5$ instead of the correct $2^5 \times 5^4$. The power of $a$ is $n-r$, the power of $bx$ is $r$; they are not interchangeable.

• C$1\,680\,000$

  This uses $\binom{9}{3} = 84$ instead of the correct $\binom{9}{4} = 126$ — a one-row-early read of the wrong binomial coefficient, with the correct powers of 2 and 5 still attached to it.

• D$20\,000$

  This is $2^5 \times 5^4$ with the binomial coefficient left out entirely — every term in the expansion needs its $\binom{n}{r}$ headcount attached, not just the two powers.

Traps tested: Power of a confused with power of bx · Term index mismatched to power · Binomial coefficient omitted

What is the coefficient of $x^3$ in the expansion of $(1 + 4x)^6$?

• $1280$

  Correct. $\binom{6}{3} = 20$, $1^3 = 1$, $4^3 = 64$, and $20 \times 64 = 1280$.

• B$80$

  This computes $\binom{6}{3} \times 4 = 80$ — the 4 raised to the power 1 instead of the power 3. $(4x)^3 = 4^3 x^3 = 64x^3$: the numeral is raised to the power just as much as the variable is.

• C$960$

  This uses $\binom{6}{2} = 15$ instead of the correct $\binom{6}{3} = 20$ — the coefficient for the WRONG term, with the correct $4^3 = 64$ still attached to it.

• D$64$

  This is $4^3$ alone, with the binomial coefficient $\binom{6}{3} = 20$ left out entirely.

Traps tested: Bx term not raised as a whole · Term index mismatched to power · Binomial coefficient omitted

What is the coefficient of $x^3$ in the expansion of $(5 - 2x)^6$?

• $-20\,000$

  Correct. $\binom{6}{3} = 20$, $5^{6-3} = 5^3 = 125$, and $(-2)^3 = -8$ since 3 is odd — so $20 \times 125 \times (-8) = -20\,000$.

• B$20\,000$

  The magnitude is right and the sign is dropped: $(-2)^3 = -8$, not $+8$. Raising a negative number to an ODD power keeps it negative.

• C$-2\,500\,000$

  This uses $5^6$ instead of the correct $5^{6-3} = 5^3$ — the power of $a$ was never reduced by $r$ at all, leaving the full power $n$ attached instead of $n-r$.

• D$-15\,000$

  This uses $\binom{6}{2} = 15$ instead of the correct $\binom{6}{3} = 20$ — the coefficient for the wrong term, with the correct powers and sign attached to it.

Traps tested: Sign lost on odd powers of a negative b · Power of a not reduced by the term index · Term index mismatched to power

Pascal's triangle is used to expand $(2 + x)^7$. Which row gives the correct coefficients, and how many entries does it have?

• Row 7 (counting the top "1" as row 0), with 8 entries: $1, 7, 21, 35, 35, 21, 7, 1$

  Correct. Row $n$ has $n+1$ entries, $\binom{n}{0}$ through $\binom{n}{n}$, and for $n=7$ that's 8 entries.

• BRow 7 (counting the top "1" as row 1), with 7 entries: $1, 6, 15, 20, 15, 6, 1$

  This is actually the row for $n=6$, read off one row too early because the count started at row 1 instead of row 0. Every coefficient taken from it is wrong, even though the additions inside the triangle were all done correctly.

• CRow 7, with 7 entries: $1, 7, 21, 35, 35, 21, 7$

  The right row, with the final entry dropped. Row 7 has $7+1=8$ entries, ending in $\binom{7}{7}=1$ — the entry for choosing bx from every single bracket.

• DRow 8, with 9 entries: $1, 8, 28, 56, 70, 56, 28, 8, 1$

  This is the row for $n=8$, one row too late — labelling the row by $n+1$ instead of $n$, the opposite direction of the more common off-by-one slip.

Traps tested: Pascal triangle row miscounted · Pascal triangle entry dropped

In the expansion of $(a+b)^n$, what does the coefficient $\binom{n}{r}$ actually count?

• The number of ways to choose which $r$ of the $n$ bracket-factors contribute a $b$ (the rest contributing $a$) when the product is expanded term by term

  Correct — this is the direct combinatorial derivation from earlier in this lesson: expanding $n$ brackets means one choice of $a$ or $b$ per bracket, and every selection choosing the same $r$ brackets for $b$ multiplies out to the same term, so the coefficient is that count.

• BThe power to which $b$ is raised in that term

  That power is $r$ itself, which is a separate quantity from the coefficient in front of the term — $\binom{n}{r}$ and $r$ are not the same number, and confusing them collapses two independent pieces of the general term into one.

• CThe total number of terms in the full expansion

  The full expansion always has $n+1$ terms in total, regardless of $r$ — that count doesn't depend on which term you're looking at, while $\binom{n}{r}$ changes with every different $r$.

• DA rounded estimate of $n$ divided by $r$

  $\binom{n}{r}$ is an exact integer count, not an estimate of anything — it's the precise number of ways to make a specific selection, computable exactly from $\dfrac{n!}{r!(n-r)!}$.

Traps tested: Binomial coefficient confused with the power · Binomial coefficient confused with total term count · Coefficient treated as an estimate not exact count

Which expression equals ${}^7C_3$?

• $\dfrac{7!}{3! \times 4!}$

  Correct: $\dfrac{7!}{3! \times 4!} = \dfrac{5040}{144} = 35$.

• B$\dfrac{7!}{3!}$

  This is $840$ — every ordered arrangement of 3 items chosen from 7, still distinguishing different orders of the same 3 items. Missing the $(7-3)! = 4!$ that removes the ordering among the 4 items left over.

• C$7 \times 6 \times 5$

  This is ${}^7P_3 = 210$, the ordered-arrangement count. Dividing by $3!$ to remove the ordering among the 3 chosen items is exactly what turns this into ${}^7C_3 = 35$.

• D$\dfrac{3! \times 4!}{7!}$

  This is the reciprocal of the correct value, $\dfrac{1}{35}$. A binomial coefficient counts a number of selections, so it's always a positive integer — never a fraction less than 1.

Traps tested: N minus r factorial omitted · Permutation used instead of combination · Ncr fraction inverted