How many numbers can be formed with the digits 1,2 3 4 3 2 1 Taken all together so that odd digits always occupy odd places?

Given:

The digits \[1,\text{ }2,\text{ }3,\text{ }4,\text{ }3,\text{ }2,\text{ }1\]

The total number of digits are \[7\]

There are \[4\] odd digits \[1,1,3,3\text{ }and\text{ }4\]odd places \[\left( 1,3,5,7 \right)\]

So, the odd digits can be arranged in odd places in \[n!/\text{ }\left( p!\text{ }\times \text{ }q!\text{ }\times \text{ }r! \right)\text{ }=\text{ }4!/\left( 2!\text{ }2! \right)\] ways.

The remaining even digits \[2,2,4\] can be arranged in \[3\]even places in \[n!/\text{ }\left( p!\text{ }\times \text{ }q!\text{ }\times \text{ }r! \right)\text{ }=\text{ }3!/2!\] Ways.

Hence, the total number of digits \[=\text{ }4!/\left( 2!\text{ }2! \right)\text{ }\times \text{ }3!/2!\]

\[=\text{ }\left[ 4\times 3\times 2\times 1\times 3\times 2\times 1 \right]\text{ }/\text{ }\left( 2!\text{ }2!\text{ }2! \right)\]

Or,

\[=\text{ }3\times 2\times 1\times 3\times 1\]

\[=\text{ }18\]

Hence, the number of ways of arranging the digits such odd digits always occupies odd places is equals to \[18\]

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How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?

There are 4 odd digits (1,3,3 and 1) that are to be arranged in 4 odd places in\[\frac{4!}{2!2!}\]ways.The remaining 3 even digits 2, 2 and 4 can be arranged in 3 even places in\[\frac{3!}{2!}\]ways.By fundamental principle of counting:

Required number of arrangements =\[\frac{4!}{2!2!}\]\[\times\]\[\frac{3!}{2!}\]= 18

Concept: Factorial N (N!) Permutations and Combinations

  Is there an error in this question or solution?

Text Solution

Solution : We have been given seven digits, namely 1, 2, 3, 4, 3, 2, 1. <br> So, we have to form 7-digit numbers, so that odd digits occupy odd places. <br> Every 7 - digits number has 4 odd places. <br> Given four odd digits are 1, 3, 3, 1 out of which 1 occurs 2 times and 3 occurs 2 times. <br> So, the number of ways to fill up 4 odd places `=(4!)/((2!)xx(2!))=6.` <br> Given three even digits are 2, 4, 2 in which 2 occurs 2 times and 4 occurs 1 time. <br> So, the number of ways to fill up 3 even places`=(3!)/(2!)=3.` <br> Hence, the required number of numbers `=(6xx3)= 18.`

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How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?

There are 4 odd digits (1,3,3 and 1) that are to be arranged in 4 odd places in\[\frac{4!}{2!2!}\]ways.The remaining 3 even digits 2, 2 and 4 can be arranged in 3 even places in\[\frac{3!}{2!}\]ways.By fundamental principle of counting:

Required number of arrangements =\[\frac{4!}{2!2!}\]\[\times\]\[\frac{3!}{2!}\]= 18

Concept: Factorial N (N!) Permutations and Combinations

  Is there an error in this question or solution?

Text Solution

Solution : We have been given seven digits, namely 1, 2, 3, 4, 3, 2, 1. <br> So, we have to form 7-digit numbers, so that odd digits occupy odd places. <br> Every 7 - digits number has 4 odd places. <br> Given four odd digits are 1, 3, 3, 1 out of which 1 occurs 2 times and 3 occurs 2 times. <br> So, the number of ways to fill up 4 odd places `=(4!)/((2!)xx(2!))=6.` <br> Given three even digits are 2, 4, 2 in which 2 occurs 2 times and 4 occurs 1 time. <br> So, the number of ways to fill up 3 even places`=(3!)/(2!)=3.` <br> Hence, the required number of numbers `=(6xx3)= 18.`

How many numbers can be formed using the digits 1,2,3,4,3,2,1 so that the odd digits always occupy the odd places?

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