Permutations and Combinations — JEE Main practice

1. Understand the setup. A strictly increasing function $f:\{1,\dots,6\}\to\{1,\dots,9\}$ is completely determined by choosing $6$ values (the images) from $\{1,\dots,9\}$ and placing them in increasing order. So counting valid functions reduces to counting valid $6$-element subsets satisfying $f(i)\neq i$.

2. Split into cases based on $f(1)$. Since $f$ is strictly increasing, once $f(1)$ is fixed large enough, all later values automatically avoid the clash $f(i)=i$.

3. Case $f(1)=2$: The remaining $5$ images are chosen from $\{3,4,\dots,9\}$ (7 values), and being placed in increasing order automatically keeps $f(i)\neq i$ for all $i$. $\binom{7}{5} = 21$

4. Case $f(1)=3$: The remaining $5$ images are chosen from $\{4,5,\dots,9\}$ (6 values). $\binom{6}{5} = 6$

5. Case $f(1)=4$: The remaining $5$ images are chosen from $\{5,6,\dots,9\}$ (5 values). $\binom{5}{5} = 1$

6. Add all the cases together. $21+6+1 = 28$

Hence, the answer is Option B (28).

1. Apply Legendre's formula for the exponent of a prime $p$ in $n!$: $\left\lfloor \frac{n}{p}\right\rfloor + \left\lfloor\frac{n}{p^2}\right\rfloor+\cdots$

2. Substitute $n=101$ and $p=7$. $\left\lfloor\frac{101}{7}\right\rfloor + \left\lfloor\frac{101}{49}\right\rfloor = 14+2 = 16$

3. Check higher powers. Since $7^3=343>101$, all higher powers contribute $0$ to the sum.

Hence, the largest value of $n$ is Option D (16).

1. Give one orange to each child first, which guarantees the 'at least one' condition. This uses up $4$ of the $16$ oranges, leaving $12$ oranges to distribute freely among the $4$ children.

2. Apply the stars-and-bars (Beggar's) method. For distributing $n$ identical items among $r$ people with no restriction, the number of ways is $\binom{n+r-1}{r-1}$. $\binom{12+4-1}{4-1} = \binom{15}{3}$

3. Compute the value. $\binom{15}{3} = \frac{15\times14\times13}{3\times2\times1} = 455$

Hence, the answer is Option D (455).