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

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- Apr 5th 2009, 02:48 AMananya janaPermutation & 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 - Apr 5th 2009, 03:36 AMPlato
- Apr 5th 2009, 03:55 AMSoroban
Hello, ananya jana!

I'll take aat what the question asks . . .*guess*

Quote:

A die is rolled 10 times.

Find the number of ways so that outcomes always contain a 1, 2, or 3.

With a sequence of 10 rolls, there are: .$\displaystyle 6^{10}$ 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: .$\displaystyle 3^{10}$ ways.

Therefore, there are: .$\displaystyle 6^{10} - 3^{19}\:=\:60,\!407,\!127$ ways.

- Apr 5th 2009, 04:13 AMananya jana
- Apr 5th 2009, 04:16 AMananya 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

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

outcomes - Apr 5th 2009, 06:52 AMSoroban
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: .$\displaystyle 6{10} = 60,\!466,\!176$ possible outcomes.

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

. . $\displaystyle \begin{array}{c|c}

\text{Case} & \text{No. of ways} \\ \hline

N1 & 5^{10} \\ N2 & 5^{10} \\ N3 & 5^{10} \end{array}$

. . $\displaystyle \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}$

. . $\displaystyle \begin{array}{c|c}

\text{Case} & \text{No. of ways} \\ \hline

N1 \cap N2 \cap N3 & 3^{10} \end{array}$

Formula: .$\displaystyle n(N1 \cup N2 \cup N3)$

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

. . $\displaystyle =\;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: .$\displaystyle 60,\!466,\!176 - 26,\!210,\!196 \:=\:\boxed{34,\!255,\!980}$ 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: .$\displaystyle 10\cdot9\cdot8 \:=\:720$ sequences of rolls.

The other seven rolls mustbe 1, 2, or 3: .$\displaystyle 3^7$ ways.__not__

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