a) How many three digit numbers can be formed from the digits 2, 3, 4, 5, 6, 7 ? [2] b) For what value of r, the coefficients of x^r and x^r+1 are equal in the expansion ( 1 + x )^11 ? [3]

a) Number of three-digit numbers from digits {2, 3, 4, 5, 6, 7}

Solution: We need to form three-digit numbers using the digits 2, 3, 4, 5, 6, 7 without repetition (since digits are distinct and no replacement is implied).

A three-digit number has:

• Hundreds place (H): 1 digit (cannot be 0, but here all digits are valid)
• Tens place (T): 1 digit (must be different from H)
• Units place (U): 1 digit (must be different from H and T)

Step-by-step calculation:

1. Hundreds place (H): Can be any of the 6 digits → 6 choices.
2. Tens place (T): Must be different from H → 5 remaining choices.
3. Units place (U): Must be different from H and T → 4 remaining choices.

Total number of three-digit numbers: $$6\times 5\times 4=120$$

Answer: 120 three-digit numbers can be formed.

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b) Value of $r$ for equal coefficients of $x^r$ and $x^{r+1}$ in $(1+x{)}^{11}$

Solution: The expansion of $(1+x{)}^{11}$ is given by the binomial theorem: $$(1+x{)}^{11}=\sum _{k=0}^{11}(\genfrac{}{}{0ex}{}{11}{k})x^k$$ Here, the coefficient of $x^k$ is $\left(\genfrac{}{}{0ex}{}{11}{k}\right)$.

We are given that the coefficients of $x^r$ and $x^{r+1}$ are equal: $$\left(\genfrac{}{}{0ex}{}{11}{r}\right)=\left(\genfrac{}{}{0ex}{}{11}{r+1}\right)$$

Using the property of binomial coefficients: $$\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\left(\genfrac{}{}{0ex}{}{n}{n-k}\right)$$ Thus, for equality: $$r=11-(r+1)$$ $$r=11-r-1$$ $$2r=10$$ $$r=5$$

Verification: Check if $\left(\genfrac{}{}{0ex}{}{11}{5}\right)=\left(\genfrac{}{}{0ex}{}{11}{6}\right)$: $$\left(\genfrac{}{}{0ex}{}{11}{5}\right)=462,\left(\genfrac{}{}{0ex}{}{11}{6}\right)=462$$ The equality holds.

Answer: The required value of $r$ is 5.