Permutation & combination problem

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April 5th 2009, 02:48 AM
ananya jana

Permutation & combination problem

A die is rolled 10 times.find th no. of ways so that outcomes always contain 1,2,3.
I cannot understand how to approach the solution.someone pls help.

THANKING YOU
April 5th 2009, 03:36 AM
Plato

Quote:

Originally Posted by ananya jana

A die is rolled 10 times.find th no. of ways so that outcomes always contain 1,2,3?

Is that the exact wording of the question?
If so, there are many different ways to read the question.
If not, please post the exact wording of the question.
April 5th 2009, 03:55 AM
Soroban

Hello, ananya jana!

I'll take a guess at what the question asks . . .

Quote:

A die is rolled 10 times.
Find the number of ways so that outcomes always contain a 1, 2, or 3.

Each roll has 6 possible outcomes.
With a sequence of 10 rolls, there are: . $6^{10}$ possible outcomes.

How many of these rolls contain no 1s, 2s, or 3s?

The rolls must consist of 4, 5, and/or 6 only.
. . There are: . $3^{10}$ ways.

Therefore, there are: . $6^{10} - 3^{19}\:=\:60,\!407,\!127$ ways.
April 5th 2009, 04:13 AM
ananya jana

Quote:

Originally Posted by Plato

Is that the exact wording of the question?

If so, there are many different ways to read the question.

If not, please post the exact wording of the question.

no the wording i have qouted as it is..
April 5th 2009, 04:16 AM
ananya jana

but we cannot count the outcome where 1 4 5 6 4 5 4 5 6 1
because 2 and 3 are absent...

but this outcome may come in the

$6^{10} - 3^{19}\:=\:60,\!407,\!127$

outcomes
April 5th 2009, 06:52 AM
Soroban

Hello again, ananya jana!

See? . . . You have a different interpretation of the problem!

If that was the exact wording, the problem is poorly written.

Quote:

A die is rolled 10 times.
Find the number of ways so that the outcomes
always contain at least one 1, one 2, and one 3.

There are: . $6{10} = 60,\!466,\!176$ possible outcomes.

Let: . $\begin{array}{ccc}N1 &=& \text{no 1's} \\ N2 &=& \text{no 2's} \\ N3 &=& \text{no 3's} \end{array}$

. . $\begin{array}{c|c} \text{Case} & \text{No. of ways} \\ \hline N1 & 5^{10} \\ N2 & 5^{10} \\ N3 & 5^{10} \end{array}$

. . $\begin{array}{c|c} \text{Case} & \text{No. of ways} \\ \hline N1 \cap N2 & 4^{10} \\ N2 \cap N3 & 4^{10} \\ N1 \cap N3 & 4^{10} \end{array}$

. . $\begin{array}{c|c} \text{Case} & \text{No. of ways} \\ \hline N1 \cap N2 \cap N3 & 3^{10} \end{array}$

Formula: . $n(N1 \cup N2 \cup N3)$
. . $=\;n(N1) + n(N2) + n(N3) -$ $\bigg[n(N1\cap N2) + n(N2\cap N3) + n(N1\cap N3)\bigg] + n(N1 \cap N2 \cap N3)$

. . $=\;3\cdot5^{10} - 3\cdot4^{10} + 3^{10} \;=\;26,210,196$

There are 26,210,196 outcomes with no 1 or no 2 or no 3.

Therefore, there are: . $60,\!466,\!176 - 26,\!210,\!196 \:=\:\boxed{34,\!255,\!980}$ outcomes
. . that contain at least one 1, one 2, and one 3.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Another interpretation . . .

Number of outcomes that contain exactly one 1, one 2, and one 3.

The 1,2,3 can occur in any of: . $10\cdot9\cdot8 \:=\:720$ sequences of rolls.

The other seven rolls must not be 1, 2, or 3: . $3^7$ ways.

Therefore, there are: . $720 \times 3^7 \:=\:\boxed{1,\!574,\!640}$ ways.