dave13 wrote:
Bunuel wrote:
A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?
(A) 144
(B) 152
(C) 160
(D) 168
(E) 176
YAY! my solution:)
\(C^3_{8}=56\) first digit cant be 0, or 1
Let's think about this one. We are not choosing 3 from 8.
You are trying to account here for . . . three slots. See below.
\(C^1_{8}=8\)
\(C^2_{9}=36*2=72\) multiply 36 by 2 since the second digit must be 0 or 1
\(C^1_{9}=9\) third digits cannot both be 0 in the same code
\(56+72+36 = 164\)
i know its wrong answer...
generis,
pushpitkc hello deep thinkers
i tried to tackle the above question using combinatorics formula but something went wrong
non of the above mentioned solutions presented in this thread feature combinatorics formula such as the one used by me...., at least i see it that way
people just use single line multiplication 8*1*10 etc ...is it another combinatorics formula that i am not aware of ? i ask this question because i dont understand based on which formula do they apply this technique ?
what`s logic
could you please correct my solution using combinatorics formula
and suggest what is name of formula such as single line multiplication
many thanks!
hope you are enjoying weekend
Hi
dave13 - combinations are unwieldy here, but
they are familiar to you. Understandable.
I am pleased to acquaint you with the Fundamental Counting Principle.
(I will PM you one way I think you can use
combinations,
pace my sanity)
• Fundamental Counting PrincipleFCP, see below, is a LOT easier.
The "line multiplication" to which you refer
is called the Fundamental Counting Principle (FCP)
FCP underlies ALL combinatorics. The posters above use it.
FCP is often called the "slot method," or more rarely, the "line method."
You MUST know about it. Bossy bear here.
For each slot, we decide how many choices we have.
____ ____ ____
Then we multiply (if there are X ways of doing Action A,
and there are Y ways of doing Action B, then there are
X * Y ways of doing Actions A AND B)
• FCP and this problemFIRST SLOT - How many of the 10 choices? No 0 or 1. Choices: 8
__8__ ____ ____
SECOND SLOT? This position is more restrictive than slot #1
(the outcome of third position is more specific, but not more restrictive)
That slot has 2 choices: slot MUST be 0
OR 1
How do we handle this restriction? In one of two ways.
(1) Split the cases
(2) Find all choices, subtract impermissible
•
TWO METHODS after the first slotMethod # 1 Split the possible cases
TWO SETS: 1 is the second digit. 0 is the second digit.
How many choices do we have for the FIRST SLOT?
10 digits total, but no 1 or 0. We have 8 choices
First slot is __8___ in BOTH cases
(1A) If "1" is in the second slot
there is ONE choice for that slot
__8__*__1__*____
Third slot? ALL 10 digits are possible
__8__ *__1__ *__10__ = 80 possible combos
(1B) If "0" is in the second slot
there is ONE choice for that slot
__8__*__1__*____
Third slot if 0 is in the second slot? How many choices?
If the second slot is 0, the third slot CANNOT be 0
We have 9 choices for 3rd slot
___8___ *___1___* ___9___ = 72 possible combos
Now what? Add or multiply or neither?
Probability rule re OR (0 OR 1, mutually exclusive) tells us
that we add the possibilities
Answer choices: too small for multiplying 72 * 80
ADD. (80 + 72) =
152 possible lock combinations
Method #2: (ALL possible) - (impermissible)
(2A) ALL, adhering to the second slot rule
ignoring (no #00) rule. Possibilities for each slot?
__8__ (no 0 or 1)
__2__ (only 0 and 1)
__10_ (pretend the third slot is not restricted yet)
__8__*__2__*__10_ = 160
(2B) IMPERMISSIBLE - How many codes WILL have #00?
Think in "MUST BE" logic
How many ways can we get a #00
Slot 2: "0" = ONE choice
Slot 3: "0" = ONE choice
__8__*__1__*__1__ = 8 impermissible cases
(ALL) - (IMPERMISSIBLE): (160 - 8) =
152Answer BHope that helps.
In addition to resources pushpitkc listed, because I know combinatorics is not taught well and often not taught at all in certain regions,
I have added many. #1 ; #2 ; #3 Bunuel 's post pf March 29, 2017 ; #4 ; #5 (please check first two links in this post); #6 ; #7 ; #8 ; #9