Waste less time on Facebook — follow Brilliant.
×

Distribution of Toys

10 different toys are to be distributed among 10 children in a way that exactly 2 children don't get a toy. Total number of ways would be?

Note: please read my approach and comment why it's wrong

Note by D K
1 year, 7 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Your approach double-counts situations where a child gets multiple toys. Suppose she gets toys A and D; she can get A as one of the eight and D as one of the two, or vice versa; you count those as two different "ways" when they are equivalent.

Mark C - 1 year, 7 months ago

Log in to reply

This is my approach( which gives the wrong answer):
First choose the 2 left outs- number of ways={10 \choose 2} ---(1)
Number of ways for 8 kids to get one toy each from 10 toys= 10X9X8...... 8-terms = {10 \choose 8} X 8! ----(2)
Number of ways to distribute remaining two toys= 8^2 -----(3)
Therefore, total number of ways to do the business= (1) X (2) X (3)

D K - 1 year, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...