How many permutations of 4 letters can be made out of the letters of the word 'examination'?
Questions of this type where permutations of a certain length are taken from a word that includes repeated letters are meant to allow repeated letters in the permutations, but identical letters are not to distinguished.For a short example, how many 3 letter permutations taken from 'ball' can be made? There are 6 with 3 distinct letters—bal, bla, abl, alb, lba, and lab—and there are 6 with 1 letter duplicated—all, lal, lla, bll, lbl, and llb. That's 12 in all.For four letter permutations of 'examination', we can consider the following cases.All four letters are different. 'examination' has 8 different letters—e, x, a, m, i, n,t , or o—so there are [math]\frac{8!}{4!}=1680[/math] of 4-permutations of these.One letter is duplicated and the other two distinct. Choose 1 of the duplicated letters—a, i, or n. There are 3 to choose from. Place them in 2 of the 4 positions. There are 6 ways of doing that. Choose 1 of the remaining 7 letters to go in the first empty position, and choose 1 of the remaining 6 letters to go in the last empty position. That makes [math]3\cdot6\cdot7\cdot6=756[/math] permutations of this typeTwo letters are duplicated. Choose 2 of the 3 doubled letters. There are 3 ways to do that. Choose 2 of the 4 positions to place the alphabetically first letter. There are [math]\binom42=6[/math] ways to do that. The other letter goes in the other two places. That makes [math]3\cdot6=18[/math] permutations of this type.Altogether, therefore, there are [math]1680+756+18=2454[/math] permutations in all.
In how many ways can 12 students be divided into 3 groups of 4 for a group project, if the order of the groups and the arrangement of the students in a group are immaterial?
12-choose-4 for the first group: 12c4 = 4958-choose-4 for the second group: 8c4 = 704-choose-4 for the third group: 4c4 = 1take the product and divide by 3! (=6) since we don't care about the order of the groups:495 * 70 * 1 / 6 = 5775
How do I calculate no. of steel bars for any beam, column, slab, and footing for given dimensions?
For beam, you will be having the size i.e width and depth. Usually, the bottom rods are important , you would have found out the area of tension steel. Minimum two rods are to be provided depending upon the width of the beam. Total area of steel DIVIDED BY 2, will give area of one rod.area of one rod is 11/14 d*d. from this you calculate the dia. This shd be nearest to 16,20,25 mm dia. If it is not suitable, provide 3 nos and follow the same procedure.For columns, mimimum no of vertical bars are to be assumed as 4. For the calculated area of steel, divide by 4, you will get area for one rod. now equate to this to 11/14 *d*d and find out d and this shd be 16,20,25 mm. If not provide six nos, repeat the above procedureFor slabs and footings, you would have calculated the spacing and from that you have to find out the number of rods EXAMPLE 8 mm dia at 200mm cto c. This means, in 1000mm [one meter], you have to provide 8 mm rods at 200mm cto c1000/200=5pacings add 1, you will get six rods per meter