# Arrange The Numbers Is There Only One Possible Arrangement I Am Desperate.

In How many ways the letters of the word RAINBOW be arranged in which vowels are never together?

-A2A-RAINBOW consists of 7 distinct letters, in which there are 3 vowels and 4 consonants.We have to find the total number of arrangements such that the vowels are never together. It could only be possible if consonants and vowels are placed alternatively.Let us arrange the  4 consonants such that there is a slot of1 l letter between ever consecutive consonants._ C _ C _ C _ C _ There are 5 alternate slots appear in the representation but we have only 3 vowels to fit in those 5 slots.3 slots from 5 empty slots can be selected in $^5C_{3} \ = 10$ ways3 vowels can be arranged in those 3 selected slots in $^3P_{3} \ = 3! \ = 6$ waysConsonants shown as C in the representation can be arranged among themselves in $^4P_{4} \ = 4! \ = 24$ waysTherefore, the total number arrangements of letters of the word RAINBOW such that vowels are never together = $\ ^5C_{3} \cdot \ ^3P_{3} \cdot \ ^4P_{4} \ = 10 * 3! * 4! \ = 1440$ waysI hope it helps!

Find the arrangement that can be made by taking 4 letters of the word MATHEMATICS?

"MATHEMATICS" As you can see there are some letters(like m,t and a) in this word which is getting repeated. So, while selecting the letters for arrangement we should consider all the cases. At first, i am gonna explain how to select letters for different cases and then later how to arrange them.We have 11 letters in "Mathematics"in which there are2 M's , 2 T's , 2 A's and other letters H,E,I,C,S are singleSelection of the 4 lettersfirst case: Two alike and other two alikeIn this case we are gonna select the two letters which are alike. We have three choices M,T,A. Out of these, we have to select two(Because we have to select four letters and selecting two alike letters means selecting four letters). So, it can be done in 3C2 ways second case: Two alike, two different1 alike letter(which will mean two letters) can be selected in 3C1 ways and other 2 different letters can be selected in 7C2 ways.(as there will be 7 different letters).So, 3C1*7C2 waysThird case: All are differentThis can be done in 8C4 ways as there are 8 different letters(M,T,A,H,E,I,C,S) Arrangement:For the first case, there will be two alike letters. So, arrangement of these letters can be done in 4!/2!*2! waysFor the second case, there will be 1 alike letter and two different letters which can be arranged in 4!/2! waysFor the third case, all letters are different so it can be arranged in 4! waysSo, the final answer will be(3C2*4!/2!*2!) + (3C1*7C2*4!/2!) + (8C4*4!) = 18+ 756 + 1680 ways= 2454 ways

Physics help! I'm desperate!?

How do I do this?

1) Suppose that a particular artillery piece has a range = 1.199×10^4 . Find its range in miles. Use the facts that 1 mile = 5280 ft and 3ft=1 yard.

2) What is the speed of a car going in SI units? Notice that you will need to change from miles to meters and from hours to seconds. You can do each conversion separately. Use the facts that 1 mile = 1609m and 1 hour = 3600 seconds

11 total letters, 4i’s, 4s’s, 2p’s, 1m.Set out 11 “slots” for each letter to go in, and then pick 4 for the i’s. From the remaining 7, choose 4 for the s’s. From the remaining 3, choose 2 for the p’s, and then finally the 1 remaining slot can take the 1m.Ie - ${11 \choose 4}{7 \choose 4}{3 \choose 2}{1 \choose 1}=\frac{11!}{7!4!}\frac{7!}{4!3!}\frac{3!}{2!1!}\frac{1!}{1!}$ which simplifies to $\frac{11!}{4!4!2!1!}=34650$.