A consider the different permutations of the word assessment...

a) Consider the different permutations of the word “ASSESSMENT” (by using each of the 10 letters exactly once) and answer the questions which follow. How many words ... i) with no restrictions? ii) begin with the letter A? iii) end with the letter T? iv) begin with the letter A and end with the letter T? v) begin with the letter A or end with the letter T? vi) begin with a vowel? vii) do not begin with a vowel? viii) have the two letters NT side-by-side? ix) have the three vowels AEE side-by-side-by-side? x) have the four letters SSSS always together? b) You are setting up a system of programming codes for a universal TV remote. Each code will have five digits, cannot include a 0 and cannot repeat any of the digits. How many codes ... i) do not contain a 5? ii) have 7 as the third digit and 9 as the fifth digit? iii) end with an even number? iv) begin with an odd number? v) end with an even number and begin with an odd number? vi) end with an even number or begin with an odd number?

Transcript

00:01 We consider the letters of the word assessment, and we want to find the number of permutations under certain constraints.

00:09 And for part one, there are no constraints.

00:14 So in assessment we have one, two, three, four, five, six, seven, eight, nine, ten letters.

00:21 However, we do have some replication.

00:23 There are four s's and two e's.

00:29 And so the number of permutations is ten factorial.

00:33 That's the number of letters.

00:33 But then we divide by 4 factorial to not count more than once identical arrangements of the s's.

00:42 And we also divide by 2 factorial to count for the two identical e's.

00:50 We get 75 ,600.

00:57 And for part 2, the number of permutations that begin with an a.

01:01 So there's only one way to fix the a at the front.

01:04 We just put the a there.

01:06 So then we can consider the remainder as being the number of permutations of the nine items.

01:11 So 9 factorial, but once again accounting for the repetitive s's and repetitive e's, 7 ,560.

01:26 For 3 we want the number of permutations that end with the letter t.

01:30 This is a similar situation to part 2 where we're fixing the single t at the end.

01:39 And so once again that's 9 factorial divided by 4 factorial times 2 factorial, 7 ,560.

01:50 Then for 4 we want the number of permutations with a at the end, or rather begin with the letter a, or end with the letter t.

02:01 So we know how many begin with a, we know how many end with t.

02:07 So there are two times this number.

02:13 But if we count all the instances of beginning with a and all the instances of ending with t, we count twice the instances that begin with a and end with t.

02:26 So we have to subtract from this the number of permutations that have a at the start, t at the end.

02:33 So one way to fix the a at the start, one way to fix the t at the end, eight remaining letters.

02:42 So we calculate the number of ways of having the a at the start and the t at the end this way.

02:51 And this is 14 ,280.

02:56 For part 5, we want the number that begin with the a or end with the letter t.

03:00 Okay, so i've made a mistake here for part four.

03:03 We only wanted the number of ways that start with the a and end with the t.

03:11 That's 840.

03:13 And then for 5, it's what i tried answering in part 4 by mistake, which is 2 times 7560 minus 840.

03:27 That was 14 ,280.

03:33 For part 6 we want the number of ways that do not begin with a vowel.

03:38 So we can see there are three vowels, the a and 2 e's.

03:45 So 7 factorial ways to arrange.

03:47 The other letters and then once again we have to count for the s's let's calculate it this way instead.

03:57 The first letter can be any of the seven letters that are not a vowel then we're arranging the remaining nine letters.

04:12 This is 52 ,920.

04:21 For part six we want the number of permutations that begin with a vowel.

04:26 So there are three ways to pick any of the three vowels and then there are nine remaining letters that we have to arrange, and then we have to count for the replication.

04:42 We get 22 ,680.

04:48 For 7 we want the number of ways that do not begin with a vowel.

04:53 That's the total number, which we calculate in part 1, minus the number of ways that do begin with a vowel, which we calculate in part 6.

05:08 52 ,920.

05:12 And then for part 8, we want the number of permutations that have the letters n and t side side.

05:21 So to do this, there's only one n and only one t.

05:24 So we can treat the two letters as a unit.

05:27 So we're finding the number of arrangements of nine items of which we have four the same and two the same of another.

05:36 But then we have to count the number of ways of arranging the n and the t within that unit.

05:40 Two ways.

05:44 This is 15 ,120...

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