Question

PLEASE DO BOTH

(1) (5 pts) How many ways can you permute the letters in BANANA? (2) (5 pts) At a Chinese restaurant, dinner for 8 people consists of 3 items from column A, 4 items from column B and 3 items from column C. If columns A, B and C have 5, 7 and 6 items respectively how many different dinner combinations are possible?

Answer 1

In this we have to use basic formulas of permutation and combinations. look at the repeated elements and divide with permutation of these also as given in images.
Que 0 the How ways can you permute many BANANA. wers in Answee Now we have 6 letters in BANANA So One letter can take 6 places , other letter can take 5 places another letter com take 4 places and So on. SO we have: One way Eways Sways 4 ways 3 Ways 2ways Eletters have 6x 5 x 4 x 3 x 2 x1 so we 6! ways but as we can see that there are 3 A's and 2 N'S we have So also counted ways N'S get inte a changed Remore that one's in Als these get which interchanged. So Our counting. fecot A's be interchange im 3X2X1 مام۸ con = 3! ways and N's 2 X ! : 2! ways So The total number of ways are: 6! 6 X 5 X 4 x 3 x 2x1 3.x2! 3 X 2 X 1 X 2 X

5 X 4 X 3 2 60 So Total the number of ways letters of BANANA can be permute are 60. for 8 of (5) At a chinese resturent dinner people Consist 3 items from column A u itters from colum Column B and 3 items from column A, B and a have 5, 7 and G items respectively, many different dinner combinations are possible. C. of how Now have take 3 items from column 5 items. A and there are to tal So we are Aso it is taking these items imidiately case of combination. total and items to be choosen. cases tak mean iie. n = Эn Such number મ - of items in qe m! (n-e)! coll for columm A choices 503 alle ll hun 56 3!(5-3)! 56 3!*2! 5X2 lo 5 X 4 X 3 X 2 l. 3X2X1 X2 XI lo choices

Now (3 for column are total 3 we choose there and 4 items xcu of items. Total choices 7! 4117-4)! 7 x 6x5x4x 3x2x/ 4x3x2 x 1X 3X2 1 = 7 X 5 35 for column B - 35 Choices. Now for column G we have to choose there are total 6 items 3 items and 6с. Total choices = 6! 31(6-3) 6X 5 X 4 X 3x2x/ 3x2x1 x 3 x 2X1 5x4 = 20 choices we see that we have to the all choose 3 from column A, u from cdumme, 10 items That m Column C 3 from so Total choices 5 Column A choices x Column Bchoices x columm a choices Total choices - 10x35 x 20 7000 Total no. different dinner combinations are = 7000