-1
$\begingroup$

In how many ways can $7$ different books be distributed to $2$ persons if each person gets at least $1$ book?

I did my calculations and my answer is $126$, but the answer stated is $216$.

$\endgroup$
2
  • 3
    $\begingroup$ How did you do the calculation? add it to your question so that we can guide you $\endgroup$
    – Heisenberg
    Jan 27, 2014 at 9:58
  • $\begingroup$ well there are 2 guys and each needs 1 book , divisions are : a. (1,6) in 7C1 * 1 ways b.(2,5) in 7C2 * 1 ways c.(3,4) in 7C3 * 1 d. (4,3) in 7C4 * 1 e. (5,2) in 7C5 * 1 and at last (6,1) in 7C6 * 1 ways and then I summed all which equals 126. $\endgroup$
    – amrx
    Jan 27, 2014 at 10:01

2 Answers 2

1
$\begingroup$

There are $7$ different books, which can each be given to one of the two people, leading to $2^7 = 128$ possible solutions. But each person should get at least one book, so the two solutions where one person gets all books should not be counted. So we get $128 - 2 = 126$.

$\endgroup$
0
$\begingroup$

Could it possibly be a typo?

If $i$ denotes the number of books person A receives then $7-i$ is the number of books person B receives. So the number of possibilities is $$\sum_{i=1}^6 \binom{7}{i}=126$$ as you calculated.

$\endgroup$
2
  • $\begingroup$ This is correct I got the same answer $\endgroup$
    – Heisenberg
    Jan 27, 2014 at 10:02
  • $\begingroup$ So I am assuming its a typo and I am correct. Moving on ...Heisenberg:) $\endgroup$
    – amrx
    Jan 27, 2014 at 10:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .