A question on permutation?

There are three unique patterns that would begin and end with odd digits:

5 * * * * 7 (middle part = permutations of 4 6 7 7, of which there are 24, 14 of which are unique.)

7 * * * * 5 (middle part = permutations of 4 6 7 7, of which there are 24, 14 of which are unique.)

7 * * * * 7 (middle part = permutations of 4 5 6 7, of which there are 24, all of which are unique.)

for the 4-digit "4 6 7 7" permutations you get the usual 4! = 1*2*3*4 = 24 combinations,

but since each of them contains two 7's, each permutation is indistinguishable from the permutation with the two 7's swapped.

That means there are half as many, since each pattern has a duplicate permutation.

14 + 14 + 24 = 48

so there are 48 unique six-digit permutations beginning and ending with an odd number here.

Considering there are 6 digits, the total number of permutations (regardless of starting or ending digits) = 6*5*4*3*2*1 = 720.

2 of the digits are even and 4 of the digits are odd. That means only 4/6 (or 2/3) of all possible permutations could start with an odd number. This narrows the permutations to 720* (2/3) = 480. However, most of these permutations would not end with an odd digit.

To further narrow down the permutations, you can already assume the first digit is odd. That means of the remaining digits, 2 of the digits are even and 3 of the digits are odd. The probability of the last digit being odd is 3/5. 480 * (3/5) = 288.

Therefore there are 288 permutations that can both start and end with an odd digit.

Another way to solve this is as follows:

1. For the first digit, there are 4 ways to fill the digit with an odd number.

2. For the last digit, one of the odd numbers is taken so there are only 3 ways to fill the digit.

3. For the 2nd digit, there are 4 remaining ways to fill it after the first and last digit are taken.

4. For the 3th digit, there are 3 remaining ways to fill it after the first 2 digits and last digit are taken.

5. For the 4th digit, there are 2 remaining ways to fill it after the first 3 digits and last digit are taken.

6. For the 5th digit, there is only 1 remaining way to fill it after all other digits are taken.

Multiply the numbers 4 * 3 * 4 * 3 * 2 * 1 = 288.

Well, you have 6 places, so ordinarily, there would be 6! combinations, but since it needs to start and end with an odd digit, you are left with only certain options.

There are 4 odd digits in the list, so the first spot can be one of 4.

The second can be anything, but there are only 5 digits left to chose from.

Third has 4 numbers left to choose from.

Fourth is has three left to choose from.

Fifth has 2 left

6th has 1 left.

You are left with 4*5*4*3*2 (*1) ... however, because the last spot must also end in an odd digit, at least two of the digits between the second and fifth slot must be even (4 and 6)

Assume one of those odds is in the 6th spot. You could have the other in either the 2, 3, 4, or 5 spot. That's 4 ways if it's the number 4 in the 6th spot, and 4 ways if it's the 6. Thus your final would be:

(4*5*4*3*2) - 8

80*6 = 480, 480 - 8 = 472

1. Number having six digits = 6!/3! = (1x2x3x4x5x6)/(1x2x3) = 720/6 = 120 ways Ans

2. Number of having 6 digits with start with odd digit and ends with odd digit = (4x 3 x4x3x2x1)/(1x2x3)

= 48 ways .....Ans

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3*(4*3*2*1)*2=144