Permutation

Hello, any help will be helpful:

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a)Find the number of even 5-digit number using digits from 2-8 with no duplication.

b) How many of the above numbers contain a 4 and if the digit 2 is there, it must come last

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Thanks in advance (Clapping)

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Originally Posted by rhhs11

a)Find the number of even 5-digit number using digits from 2-8 with no duplication.

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i suppose 2 and 8 are included here.

to be even, the last digit must be 2,4,6 or 8. so there are 4 choices for the last digit. for the next digit, we have 6 choices (because there are 6 digits left), for the next, there are 5 choices (because we cannot choose something we already chose). continuing like this, we have the answer to be:

4(6)(5)(4)(3) = 1440

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b) How many of the above numbers contain a 4 and if the digit 2 is there, it must come last

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clarification here: what must come last?

See if I count these right.

a): Since the number is even, it must end in 2,4,6,8

There are then 4 choices for the ending digit, but the other three can be distributed among the other spaces. Since there is not repetition, we have 6 choices for the fourth, 5 for the third, and so forth.

4(3*4*5*6)=1440


b): Place the 2 in thr last spot. Then the 4 can go in any of four spaces.
There are 60 choices for the other 3 spots. 4*60=240.

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Originally Posted by Jhevon

clarification here: what must come last?

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2 must come last. What i did was that i got the 1440 - (must have 4)[1*6*5*4*3) = 1080. Then, from that 1080, i did 1080 - (1*1*6*5*4) = 960.

The answer is 552!!

Thanks for the replies, but please help!