In how many ways the letters of word simple can be arranged so that all vowels come together

A fraction is 0 if and only if​

Denise’s rabbit can eat 70 kilos of food in 80 days. How long will it take the rabbit to eat 87. 5 kilos?.

pa tulong po sa pag answer ​

Given the function f[x]=x²–4x , Find a. f[2] b. f[a+b] c. f[2–x]​

The common ratio of 64 32 16 8

The common ratio of this GPs 64 32 16 8

Mang juan harvested coconut fruits in their farm to make wine [tuba]. He filled his big jar with 10 12 gallons of coconut fruit juice. He consumed 4 1 … 4 gallons and gave 3 14 gallons to his friend. How many gallons of coconut fruit juice did mang juan has left?.

help plss with solutions... no nonsense pls Q A Q

Mr. Jo is a known teacher in area of Mathematics. He is very much respected in the discipline. However, his knowledge does not go beyond it. Engage hi … m in other discipline and he would not be able to give the same depth as when he talks about Mathematics. As a teacher, does Mr. Jo need to know other discipline? If yes what particular teachersâ role is the knowledge of other disciplines related to? Why?.

plss help no nonsense pls with solution

answer with explanation

Answer: Option A

Explanation:

Number of ways of selecting 3 consonants from 7
= 7C3Number of ways of selecting 2 vowels from 4

= 4C2

= 7C3 × 4C2

It means we can have 210 groups where each group contains total 5 letters [3 consonants and 2 vowels].

$=210×120=25200$

answer with explanation

Answer: Option B

Explanation:

In a group of 6 boys and 4 girls, four children are to be selected such that at least one boy should be there.Hence we have 4 options as given below

We can select 4 boys ...[option 1]

We can select 3 boys and 1 girl ...[option 2]

We can select 2 boys and 2 girls ...[option 3]

We can select 1 boy and 3 girls ...[option 4]

= 6C4 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C3

$=\dfrac{6×5}{2×1}+\dfrac{6×5×4}{3×2×1}×4$ $+\dfrac{6×5}{2×1}×\dfrac{4×3}{2×1}+6×4$

answer with explanation

Answer: Option C

Explanation:

From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.Hence we have the following 3 options.We can select 5 men ...[option 1]

Number of ways to do this = 7C5

Number of ways to do this = 7C4 × 6C1

Number of ways to do this = 7C3 × 6C2

= 7C5 + [7C4 × 6C1] + [7C3 × 6C2]

$= \dfrac{7×6}{2×1}+\dfrac{7×6×5}{3×2×1}×6$ $+\dfrac{7×6×5}{3×2×1}×\dfrac{6×5}{2×1}$

answer with explanation

Answer: Option C

Explanation:

The word 'OPTICAL' has 7 letters. It has the vowels 'O','I','A' in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is, PTCL[OIA].Hence we can assume total letters as 5 and all these letters are different.Number of ways to arrange these letters$=5!=5×4×3×2×1=120$All the 3 vowels [OIA] are differentNumber of ways to arrange these vowels among themselves$=3!=3×2×1=6$Hence, required number of ways

$=120×6=720$

answer with explanation

Answer: Option C

Explanation:

The word 'CORPORATION' has 11 letters. It has the vowels 'O','O','A','I','O' in it and these 5 vowels should always come together. Hence these 5 vowels can be grouped and considered as a single letter. that is, CRPRTN[OOAIO].Hence we can assume total letters as 7. But in these 7 letters, 'R' occurs 2 times and rest of the letters are different.Number of ways to arrange these letters

$=\dfrac{7!}{2!}=\dfrac{7×6×5×4×3×2×1}{2×1}=2520$

find n if np2=72

np2=72 n[n-1] = 72n^2-n-72=0[n-9][n+8]=0

n = 9

if teams A, B, C, D, E, F, G, H, I, J, K, L, M, are having a competition, how many ways would their result be represented. please solve  by use of  a tree diagram.

how many ways can 13 football teams display their results, please indicate the pattern of possible results

Assuming that the possible results each team can have are SUCCESS, FAILURE and DRAW and all teams need to display any of the result.

Total number of results possible = 3^13

In how many ways can the letters of the word UNIQUE be permitted such that the vowels are in even places? 

vowels can be arranged in 3! ways=6

and remaining letter can be arranged in 3!=6

total=6*6=36 ways

No arrangements possible. vowels should be  in even places. Could you show at least one such arrangement?

Why is it impossible Neeraj...... Vowels should be in even places means it should be either in 2nd, 4th or 6th place in the word. So the answer given by Santosh is correct. If you want to show you one word of that kind- 'UINUQE'. Don't be in confusion the word which will be formed must have dictionary meaning. If nothing has been mentioned then by default we have to take the no of arrangements possible without having dictionary meaning.

@Amlan, the word you have given is 'UINUQE'. In that, 'U' comes in 1st position. But vowels are allowed to come in even places only.  So I guess Neeraj is right.

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12 Qs. 36 Marks 20 Mins

Calculations:

Number of alphabets in word TODAY = 5

Number of arrangements of word TODAY = 5!

⇒ Number of arrangements of word TODAY = 5 × 4 × 3 × 2 × 1

⇒ Number of arrangements of word TODAY = 120      ----(1)

Arrangement with vowels together = 4! = 24

Number of arrangements of vowels amongst themselves = 2! = 2

Number of arrangements with vowels together = 24 × 2 = 48      ----(2)

Required arrangements = Total arrangement - Arrangement with vowels together

⇒ Required number of arrangements = 120 - 48     [From (1) and (2)]

⇒ Required number of arrangements = 72

∴ The required number of arrangements is 72.

Additional Information

The number of ways of arranging unlike letters of an 'n' lettered word = n!

n! = n × (n - 1) × (n - 2) ×........× 1

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