🔒 Closed Math questions and problem

[0MeGaMinD0]

Sa lahat ng ka PHC pwede po patulong tungkol sa permutation: How many 3-digit numbers can be formed from the digits 1,3,4 and 6 if each digit can be used only once?, para po sa aming report.....Salamat

 

[bagong pangalan]

sakit ng ulo tingnan mga ganitong math

 

[Jmrie_]

eto read mo

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[Norton360]

24 ang sagot
4 x 3 x 2 x 1

 

[Norton360]

4x3x2 lang pala

 

[0MeGaMinD0]

4x3x2 lang pala

pwede po ba akong bigyan ng complete solution?sir?

 

[0MeGaMinD0]

4x3x2 lang pala

Mayroon po kayong complete solution?...bigay po kayo

 

[0MeGaMinD0]

Ano naman kung ganito ang problem?: How many 3-digit odd numbers can be formed from the digits 2,5,7,8 and 9 if each digit can be used only once?

 

[maudesune]

Okay. Uhm.

1. How many 3-digit odd numbers can be formed from the digits 2, 5, 7, 8, and 9?

Since there are 5 digits, the solution is basically
5! divided by 2!

= 5 x 4 x 3
= 60

2. How many of them are odd numbers?

So basically, we need to get the numbers that end in 5, 7, and 9.
So the odd numbers ending in 5, for example, are the permutation of the remaining 4 digits.
(We are trying to find 2-digit numbers that can be formed from the rest of the digits.)

That's
4! divided by 2!

= 4 x 3
= 12

But since there are 3 odd numbers

12 x 3

= 36 total odd numbers


I hope it's coherent enough.
orz

Edit: I'll try to construct an easier way of understanding this sort of math problems. Pero sa pag-uwi ko na. u-u)b

 

[0MeGaMinD0]

Okay. Uhm.

1. How many 3-digit odd numbers can be formed from the digits 2, 5, 7, 8, and 9?

Since there are 5 digits, the solution is basically
5! divided by 2!

= 5 x 4 x 3
= 60

2. How many of them are odd numbers?

So basically, we need to get the numbers that end in 5, 7, and 9.
So the odd numbers ending in 5, for example, are the permutation of the remaining 4 digits.
(We are trying to find 2-digit numbers that can be formed from the rest of the digits.)

That's
4! divided by 2!

= 4 x 3
= 12

But since there are 3 odd numbers

12 x 3

= 36 total odd numbers


I hope it's coherent enough.
orz

Edit: I'll try to construct an easier way of understanding this sort of math problems. Pero sa pag-uwi ko na. u-u)b

Paano naman yung post ko na: How many 3-digit numbers can be formed from the digits 1,3,4 and 6 if each digit can be used only once? Bigyan mo po ng solution boss Math master

 

[maudesune]

How many 3-digit numbers can be formed from the digits 1,3,4 and 6 if each digit can be used only once?

There are four digits so--

4! divided by 1!

= 4 x 3 x 2
= 24

 

[maudesune]

How to solve permutations:

Problem example:

How many 3-letter combinations can be formed from the letters A, B, C, D, E, F?

1. Count the number of letters. A B C D E F = 6 letters.
So that means

6! (6 factorial) = 6 x 5 x 4 x 3 x 2 x 1

2. Determine the number of combination.

Yung given is 3-letter. So that means

3! (3 factorial) = 3 x 2 x 1

3. Divide them. (Alternate on #4)

6! divided by 3! = 6 x 5 x 4

4. Easier way of doing it.

6! = 6 x 5 x 4 x 3 x 2 x 1

Tapos 3-letter combination di ba?

So count ka lang

6! = 6 x 5 x 4 x 3 x 2 x 1

Bale yung tatlo lang kukunin mo.

= 6 x 5 x 4
= 120 possible combinations.

 

[Norton360]

pwede po ba akong bigyan ng complete solution?sir?

solution na yun, bale sabi mo how many 3 digit numbers can be formed by digits 1, 3, 4 and 6 without repetitions. bale may 4 ka na numbers, pro 3 digit numbers lang w/o repetitions kailangan mo, gawa ka ng 3 blanks para sa 3 digit no. mo, each blank represents a single digit, sa first blank, makakapili ka ng 4 na digit, sa 2nd blank may 3 digits ka nalang kasi napili na ang isa, at sa huli may 2 nalang kasi nakapili ka na ng dalawa, so 4x3x2