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how many 2 digit numbers can be formed from 1,1,2,3? the ans is (11,12,13,21,23,32,31) can any one explain how to solve this using permutations??please..... similarly I have 2 red,1green & 1 blue ,how many ways I can select 2 out of those? here also ans is (RR,RB,RG,BG) please explain this using combinations formula? I'm unable to do this

Ajay shifu
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3 Answers3

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In both problems, you have 3 types (2 of one type, and 1 each of some types)

You can simplify into two cases (a) 2 selected of one type (b) 2 selected of 2 types.

For the first problem,

$\text {use the formula} : {2\choose 2} + {3\choose 2}\cdot2!$

$\text {or, since you mentioned permutations} : {\dfrac{^2P_2}{2!}} + ^3P_2$

Do the 2nd problem in similar fashion, only using combinations : ${2\choose 2} + {3\choose 2}$

true blue anil
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  • tq so much...but still have confusion...so answer another question so that I'll get a clarity....how many 2 digit numbers can be formed from 1,1,2,2? – Ajay shifu Aug 11 '15 at 06:18
  • $2\times ^2P_2/2! + ^2P_2$ – true blue anil Aug 11 '15 at 06:21
  • how did u get dat? I mean I'm really weak in dis....I can't understand.... how u r writing dat 2x 2p2 /2! + 2p2 sorry I'm really weak – Ajay shifu Aug 11 '15 at 06:27
  • @Ajay Right, looks like you are asking for a general solution. I have edited the question, please accept if that is what you asked – Pagol Aug 11 '15 at 06:41
  • Looking only at the first term, you can select "doubles" in 2 ways (either two 1's or two 2's) that is why the multiplication by 2 and need to divide the permutations by 2! because both the digits are identical. – true blue anil Aug 11 '15 at 06:51
  • In my opinion, the right question would be: if we have $n$ objects with clusters $1, 2, \dots, c$ where there are $n_1, n_2, \ldots, n_c$ identical objects, with $\sum_{i=1}^c n_i = n$, how many permutations of length $m$ are possible?

    Similarly, if we have 2 red, 1 green & 1 blue ball, in how many ways can we select 2 out of those? Here the answer is (RR,RB,RG,BG), and an equivalent exercise can be repeated when ordering of the elements does not matter.

     – Pagol Aug 11 '15 at 06:57
  • @Pagol: Of course, a general formulation is good, but do you really think that the OP is in a position to understand a more complex formulation, when struggling with the simpler version ? – true blue anil Aug 11 '15 at 07:04
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The idea is to enumerate how many ways you can find $(k_1,\ldots,k_c) =: k$ such that $0 \leq k_i \leq n_i, \forall i = 1, \ldots, c$ and $\sum_{i=1}^c k_i = m$. For your original $c = 3$ clusters and $m = 2$ element picking problem, there are 4 such $k$'s: $(0,1,1), (1,0,1), (1,1,0), (2,0,0)$, clustering the first two digits ((1,1), (2), (3)). Once you have this formulation, the answer is really easy:

Let $\kappa(n,m) := \{k \in \mathbb{Z}_+^c : 0 \leq k_i \leq n_i, \forall i = 1, \ldots, c, \sum_{i=1}^c k_i = m\}$.

The answer is: $$\sum_{k \in \kappa(n,m)} \frac{m!}{\prod_{i=1}^c (k_i!)}.$$

Plug the values in your case: $\frac{2!}{0!1!1!} + \frac{2!}{1!0!1!} + \frac{2!}{1!1!0!} + \frac{2!}{2!0!0!} = 7$.

For the combinations: just divide it in each stage by the number of repetitions. Define $p(k) = |\{i : k_i > 0\}|$. The answer is: $$\sum_{k \in \kappa(n,m)} \frac{1}{p(k)} \frac{m!}{\prod_{i=1}^c (k_i!)}.$$

Plug values in your case: $\frac{1}{2} \frac{2!}{0!1!1!} + \frac{1}{2} \frac{2!}{1!0!1!} + \frac{1}{2} \frac{2!}{1!1!0!} + \frac{1}{1} \frac{2!}{2!0!0!} = 4$.

Pagol
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Approach via cases:

Case 1: The first digit is a $1$.

How many choices are there for the second digit?

It will be any of $1,2$ or $3$, for a total of three possibilities

Case 2: The first digit is not a $1$.

How many choices are there for the first digit?

It will be either a $2$ or a $3$, for a total of $2$ possibilities for this step

How many choices are there for the second digit after having picked the first?

It can't be whatever you picked for the previous step, so it is either a $1$ or whichever of $2$ or $3$ wasn't used, for a total of $2$ possibilities for this step.

Apply the multiplication principle and the addition principle to reach the final conclusion. This problem does not use permutations nor does it use combinations. It uses more fundamental concepts that should have been learned before either permutations or combinations were introduced.

This yields a total answer of $3 + 2\cdot 2 = 7$ possibilities ($3$ from case 1, $2\cdot 2$ from case 2) matching the answer given.

For the second question where order of selection is unimportant, you can again approach via cases.

Case 1: There is a $R$

What is the other color?

It can be $R,G$ or $B$ for a total of three possibilities

Case 2: There is not a $R$

If you are always picking two colors and you aren't allowed to pick $R$, then there is only one way to pick them since only two remain.

Again, there was no need here to approach via combinations.

The final answer yields a total of $3+1$ possibilities ($3$ from case 1, $1$ from case 2).

JMoravitz
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