in how many ways the letter of the word STRANCE be arranged so that the vowels occupy only the odd place

1440

Step-by-step explanation:

We're given the word ‘STRANCE’ STRANCE happens to be a 7 letterered word. It contains just 2 vowels A and E, with 5 consonants S, T, R, N and C. We're told to arrange in a way that the vowels occupy only the odd places, so in the word, there exist 4 odd places. The 1st, 3rd, 5th and 7th spot.

The two vowels can be arranged in 4P2 ways. This means that the remaining 5 consonants can be arranged among themselves in a 5P5 ways. Using the permutation formula, we know that

P (n, r) = n!/(n – r)!, so

P (4, 2) * P (5, 5) =

4!/(4 – 2)! * 5!/(5 – 5)! =

4!/2! * 5! =

(4 * 3 * 2!)/2! * 5! =

(12 * 2!)/2! * 120 =

(12 * 2)/2 * 120 =

12 * 120 = 1440

Ok You're probably here from K12 so heres the

Conclusion: If a polygon undergoes a rigid transformation, then the image and pre-image have equal perimeters.

I just took the test and got this right, I'm a K12 student as well.

Hope this helps :)