Testbook

Question

In how many ways 10 different books can be distributed among there students A, B and C such that A must be given at least 1 book?

Answer 1

There are $\binom{10}{1}$ ways to select one book for student A. Choosing which book among the remaining is given to student B determines which books go to student C. For student B, we can select 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 of the remaining 9 books. Hence, the number of arrangements in which one book is given to student A are:

$\binom{10}{1} [\binom{9}{0} + \binom{9}{1} + \binom{9}{2} + ...+\binom{9}{9}] = \binom{10}{1}2^{9}$

There are $\binom{10}{2}$ ways to select two of the 10 books to be given to student A. For student B, we can select 0, 1, 2, 3, 4, 5, 6, 7, 8 of the remaining 8 books. Hence, the number of arrangements in which exactly two people are given to student A are:

$\binom{10}{2} [\binom{8}{0} + \binom{8}{1} + \binom{8}{2} + ...+\binom{8}{8}] = \binom{10}{2}2^{8}$

Continuing in this way, we obtain the total number of arrangements as:

$\binom{10}{1}2^{9} + \binom{10}{2}2^{8} + \binom{10}{3}2^{7} + ...\binom{10}{10}2^{0}$

I don't know if there's a way to simplify this expression. This gives the answer 58025. http://ideone.com/rPYmKy

Comments

$$\begin{align*} &\;\;\; \;\; \text{standard result} \\ &\Rightarrow \sum_{k=0}^{n}2^k.\binom{n}{k} = 3^n \\ \\ \hline \\ &\text{Now, what you have arrived till now} \\ &= \sum_{k=0}^{9}.\binom{10}{k}.2^{k} \\ & \text{do following ,} \\ &= \sum_{k=0}^{9}.\binom{10}{k}.2^{k} + \binom{10}{10}.2^{10} - \binom{10}{10}.2^{10} \\ &= \sum_{k=0}^{10}.\binom{10}{k}.2^{k} - \binom{10}{10}.2^{10} \\ &= 3^{10} - 2^{10} = 58025 \\ \end{align*}$$

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Above standard result $\rightarrow$

$$\begin{align*} \color{green}{\sum_{k=0}^{n} \left [ \binom{n}{k}.2^{k}.1^{n-k} \right ] = \left ( 1+2 \right )^n =\sum_{k=0}^{n}\binom{n}{k}.2^{k}} \end{align*}$$

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#air1..you know how to simplify, because it involves only binomial formula but rather you relied on python ncr. But nice method. (y)

Answer 2

use complementary counting .

58025 is correct answer .