Question

= 66 - ($^{17}C_{11} - 6^{11}C_5$).

ILLUSTRATION 7.127

In how many ways can 14 identical toys be distributed among three boys so that each one gets at least one toy and no two boys get equal number of toys.

Answer 1

-9538

Answer 2

The problem asks to evaluate the given expression: $= 66 - ($^{17}C_{11} - 6^{11}C_5$).

We need to calculate the values of the combination terms first. The formula for combinations is $^nC_r = \frac{n!}{r!(n-r)!}$.

Step 1: Calculate $^{17}C_{11}$. Using the property $^nC_r = ^nC_{n-r}$, we have $^{17}C_{11} = ^{17}C_{17-11} = ^{17}C_6$. $^{17}C_6 = \frac{17!}{6!(17-6)!} = \frac{17!}{6!11!} = \frac{17 \times 16 \times 15 \times 14 \times 13 \times 12}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$ Simplify the expression: $^{17}C_6 = 17 \times \frac{16}{4} \times \frac{15}{5 \times 3} \times \frac{14}{1} \times \frac{13}{1} \times \frac{12}{6 \times 2 \times 1}$ $^{17}C_6 = 17 \times 4 \times 1 \times 14 \times 13 \times 1$ $^{17}C_6 = 68 \times 182$ $^{17}C_6 = 12376$

Step 2: Calculate $^{11}C_5$. $^{11}C_5 = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1}$ Simplify the expression: $^{11}C_5 = 11 \times \frac{10}{5 \times 2} \times \frac{9}{3} \times \frac{8}{4} \times 7$ $^{11}C_5 = 11 \times 1 \times 3 \times 2 \times 7$ $^{11}C_5 = 11 \times 42$ $^{11}C_5 = 462$

Step 3: Calculate $6 \times ^{11}C_5$. $6 \times 462 = 2772$

Step 4: Substitute the values back into the original expression. $66 - ($^{17}C_{11} - 6^{11}C_5$) = 66 - (12376 - 2772)$ First, evaluate the term inside the parenthesis: $12376 - 2772 = 9604$

Now, complete the final subtraction: $66 - 9604 = -9538$

The final value of the expression is -9538.

The illustration provided in the question text is unrelated to the calculation required.