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How many 4-digit even numbers can be formed using digits 1,2,3,5 using each digit once? i cant understand how to solve it...

and please also define Find the number of words that can be formed from the word VIRTUAL if the words end with letter T and start with letter L. (Words need not be in the dictionary and each letter should be used once)

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    $\begingroup$ It must end in 2. The other digits have to be 1,3, and 5. How many was are there to arrange the numbers 1,3, and 5? Think harder. $\endgroup$ – fleablood Jan 21 '17 at 22:12
  • $\begingroup$ I think I saw this question a few days ago $\endgroup$ – Travis Jan 21 '17 at 22:16
  • $\begingroup$ no it is not because i have searched every where $\endgroup$ – rajput Jan 21 '17 at 22:41
  • $\begingroup$ welcome to maths stack exchange. do not add new question to your original question after others have answered your questions. most answers end up not answering the new question completely. Also, show us your thoughts and attempts to help us help you better. $\endgroup$ – Siong Thye Goh Jan 21 '17 at 23:03
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$2$ must be the last digit because $1, 2$, and $3$ are not divisible by 2, and not even. So the simple rules of how many ways we can arrange things comes into place. Because there are 3 other digits, the way we can arrange the 4 digits $2$ being the last one is $3!$.

$3! = 6$, 6 4 digit even numbers.

The reason $3!$ works is because $3! = 3\times 2 \times 1$, and we have 3 digits, $1,2,3$. All 3 numbers can always be the first number. If one number is first, there are 2 numbers left to be the second digit, and if 2 numbers are digits 1 and 2, only one number is left to be the last digit, hence $3 \times 2 \times 1$

This property applies to any chain of $n$ length. The amount of ways we cabn arrrange the chain $= n \times (n-1) \times (n-2) \ \ldots \times 2 \times 1$, or $n!$

As this applies to $VIRTUAL$, it can be expressed as a chain of length 5 because the first and last letters never change. $5! = 120$

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Hint:

The only way it could be even is if it ended with $2$.

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Hint:

Let the number that you want to construct be $abcd$

pick a digit for $d$. how many choices do you have?

Now, pick a digit for $c$ after you pick $d$, how many choices do you have?

Now, pick a digit for $b$ after you pick $c$ and $d$, how many choices do you have?

Now, pick a digit for $a$ after you pick $b, c$ and $d$, how many choices do you have?

Multiply them up.

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  • $\begingroup$ if i multiplay 4*3*2*1 dcba then answer is wrong $\endgroup$ – rajput Jan 21 '17 at 22:33
  • $\begingroup$ $d$ must be $2$, you only have one choice for $d$. $1 \times 3 \times 2 \times 1$. $\endgroup$ – Siong Thye Goh Jan 21 '17 at 22:46
  • $\begingroup$ if i have one choice then how i chose next ??? $\endgroup$ – rajput Jan 21 '17 at 22:48
  • $\begingroup$ $1$ choice for $d$ but you have $3$ choices for $c$ right? $\endgroup$ – Siong Thye Goh Jan 21 '17 at 22:58
  • $\begingroup$ yes i understand $\endgroup$ – rajput Jan 21 '17 at 23:11
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This numbers always ends with 2. So answer is count of permutations for $\{1 , 3 , 5\}$ = $3!$ = $6$

Second problem: First and last characters are fixed. Then use idea from first problem

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