Permutations & Combinations
Permutations & Combinations
star_batch_jee_advanced_2025
Grade 11
Question:
There are counters available in $3$ different colours (atleast four of each colour). Counters are all alike except for the colour. If '$m$' denotes the number of arrangements of four counters if no arrangement consists of counters of same colour and '$n$' denotes the corresponding figure when every arrangement consists of counters of each colour, then:
$m = 2n$
$6m = 13n$
$3m = 5n$
$5m = 3n$
Step-by-Step Solution
Key Concept: Differentiating the binomial identity and integrating term-by-term creates a sum of rational expressions involving binomial coefficients.
Starting with the binomial expansion $\sum_r ^nC_r(1-y)^r(-1)^r y^{n-r} = x^n$, differentiate with respect to $y$ to get $\sum_r ^nC_r(-1)^{r-1}(1-y)^{r-1} - \frac{1-x^n}{1-x}$. Integrate from a limit and rearrange to obtain $\sum_{r=1}^{n} \frac{^nC_r(1-y)^r(-1)^{r-1}}{r} = 1-x + \frac{1-x^2}{2} + \frac{1-x^3}{3} + ... + \frac{1-x^n}{n}$.
Correct Answer: 2