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CSAT 2023

Q19 2023, Permutation and Combination

How many distinct 8-digit numbers can be formed by rearranging the digits of the number 11223344 such that odd digits occupy odd positions and even digits occupy even positions?

संख्या 11223344 के अंकों को पुनर्व्यवस्थित कर भिन्न 8-अंकों की कितनी संख्याएँ बनाई जा सकती हैं, इस प्रकार कि विषम अंक विषम स्थानों पर हों और सम अंक सम स्थानों पर हों?

Explanation Please wait..

In an eight-digit number, there are 4 positions for odd numbers and 4 positions for even numbers.

The number 11223344 has 4 odd numbers (1, 1, 3, 3) and 4 even numbers (2, 2, 4, 4).

Since we're arranging these digits such that odd digits are in odd positions (1st, 3rd, 5th, 7th) and even digits are in even positions (2nd, 4th, 6th, 8th), we have the following pattern:

OEOEOEOE

Where O represents an odd digit and E represents an even digit.

Step 1: First, let's count the number of ways to arrange the odd digits in the odd positions.

Since there are two 1's and two 3's, we treat them as distinct when arranging them.

The number of ways to arrange 1, 1, 3, 3 in the odd positions can be calculated using the formula for permutations of objects with repetitions.

It is given by

Step 2: Next, we count the number of ways to arrange the even digits in the even positions.

Similarly, since there are two 2's and two 4's,

we have ways to arrange them.

Step 3: Multiplying these two possibilities together gives us the total number of distinct 8-digit numbers that can be formed:

Total = 6 (arrangements of odd digits) * 6 (arrangements of even digits) = 36

So, number of such distinct numbers = 6 × 6 = 36

Thus, option (c) is the correct answer.

In an eight-digit number, there are 4 positions for odd
numbers and 4 positions for even numbers.

The number 11223344 has 4 odd numbers (1, 1, 3, 3) and 4
even numbers (2, 2, 4, 4).

Since we're arranging these digits
such that odd digits are in odd positions (1st, 3rd, 5th, 7th) and even digits
are in even positions (2nd, 4th, 6th, 8th), we have the following pattern:

OEOEOEOE

Where O represents an odd digit and
E represents an even digit.

Step 1: First,
let's count the number of ways to arrange the odd digits in the odd positions.

<ul><li> 
Since
there are two 1's and two 3's, we treat them as distinct when arranging them.</li></ul>

<ul><li> 
The
number of ways to arrange 1, 1, 3, 3 in the odd positions can be calculated
using the formula for permutations of objects with repetitions.</li></ul>

<ul><li> 
It
is given by <v:shapetype id="_x0000_t75" coordsize="21600,21600" o:spt="75" o:preferrelative="t" path="m@4@5l@4@11@9@11@9@5xe" filled="f" stroked="f">
 <v:stroke join="">
 <v:formulas>
 <v:f eqn="if lineDrawn pixelLineWidth 0">
 <v:f eqn="sum @0 1 0">
 <v:f eqn="sum 0 0 @1">
 <v:f eqn="prod @2 1 2">
 <v:f eqn="prod @3 21600 pixelWidth">
 <v:f eqn="prod @3 21600 pixelHeight">
 <v:f eqn="sum @0 0 1">
 <v:f eqn="prod @6 1 2">
 <v:f eqn="prod @7 21600 pixelWidth">
 <v:f eqn="sum @8 21600 0">
 <v:f eqn="prod @7 21600 pixelHeight">
 <v:f eqn="sum @10 21600 0">
 </v:f></v:f></v:f></v:f></v:f></v:f></v:f></v:f></v:f></v:f></v:f></v:f></v:formulas>
 <v:path o:extrusionok="f" gradientshapeok="t" o:connecttype="rect">
 </v:path></v:stroke></v:shapetype><v:shape id="_x0000_i1025" type="#_x0000_t75">
 <v:imagedata src="file:///C:\Users\GAURAV~1\AppData\Local\Temp\msohtmlclip1\01\clip_image001.png" o:title="" chromakey="white">
</v:imagedata></v:shape></li></ul>

Step 2: Next,
we count the number of ways to arrange the even digits in the even positions.

<ul><li> 
Similarly,
since there are two 2's and two 4's,</li></ul>

<ul><li> 
we
have <v:shape id="_x0000_i1025" type="#_x0000_t75">
 <v:imagedata src="file:///C:\Users\GAURAV~1\AppData\Local\Temp\msohtmlclip1\01\clip_image001.png" o:title="" chromakey="white">
</v:imagedata></v:shape> ways to arrange them.</li></ul>

Step 3: Multiplying
these two possibilities together gives us the total number of distinct 8-digit
numbers that can be formed:

<ul><li> 
Total
= 6 (arrangements of odd digits) * 6 (arrangements of even digits) = 36</li></ul>

<ul><li> 
So,
number of such distinct numbers = 6 × 6 = 36</li></ul>

<ul><li> 
Thus,
option (c) is the correct answer.</li></ul>

CSAT Year Wise (Temporary Link, This will be removed and no PDFs shall be downloaded here. Please download CSAT subject wise files) / सी सैट

Q19 2023, Permutation and Combination

How many distinct 8-digit numbers can be formed by rearranging the digits of the number 11223344 such that odd digits occupy odd positions and even digits occupy even positions?

Explanation Please wait..

व्याख्या Please wait..

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John Doe

CSAT Year Wise (Temporary Link, This will be removed and no PDFs shall be downloaded here. Please download CSAT subject wise files) / सी सैट

Q19 2023, Permutation and Combination

How many distinct 8-digit numbers can be formed by rearranging the digits of the number 11223344 such that odd digits occupy odd positions and even digits occupy even positions?

Explanation Please wait..

व्याख्या Please wait..

Quick Links

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