please answer a ques …8 students qualify for final. all but 2 will advance to finals and out of those only 3 will get mealdals. findout how many such groups of 3 are possible.
wow !!!u rock who evr u r thanx can u pls help me trig 2 pls pls with sugar in top pls
It was really a fabulous explanation 4 Permutation & Combination.Now I’m feeling much relaxed after deliberately going through this…
please let me know why you multiply the number of choices when doing the permutation. Why not addition?
Ignore my previous comment. The email address was not right there.
Please explain me why you multiply the choices when doing permutation, not addition, or subtraction, or division?
You can definitely see your enthusiasm in the work you write. The sector hopes for more passionate writers such as you who aren’t afraid to mention how they believe. Always follow your heart.
sir, i did a lot of question by the lesson PERMUTION AND COMBINATION but one thing i don’t have any idea to solve about this question,
( LCM)- 4!,5!,6!
I request you to understand me about it.
Hi Sabra, great comment, thanks for sharing your think process. I agree, the language can be confusing/counter intuitive. Here’s another take: with permutations, shuffling counts as a new arrangement, with combinations, shuffling does not count.
It’s more important to find the wording that clicks with you vs. the standard “order matters” language which, I must admit, is not immediately clear. “Shuffles count as new items” or similar may click better!
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i already printed the comments up to post # 667… it helps me a lot, as a BSED-MATH STUDENT… ty so much… but, im still confused on the four letter combination from the word OUTLOOK… pls help me… especially kalid… :))
i already printed the comments up to post # 667… it helps me a lot, as a BSED-MATH STUDENT… ty so much… but, im still confused on the four letter combination from the word OUTLOOK… pls help me… especially kalid
dam gud explanation…wow…
@Shabazz: Glad it helped!
Sorry for the double post, but my post came out very poorly from my mobile device. So here is it clean from my desktop.
I am calculating the probability of major earthquake occurrence rates.
over the last 112 years 151 major earthquakes occurred M7 or greater. That gives us an overall 151/1335 chance of occurrence in any given month.
p= 0.11310861423221 = 151/1335 chance of occurrence per month
t= number of months over which to find probability of occurrence of n number of occurrences.
n= number of occurrences
I am currently working with this.
probability of exactly n quakes (no more no fewer) occurring in t months if p is the overall probability for one month =
= p^t * (1-p)^(t-n) * the number of permutations of n occurrences in t times
Currently, I find the factor programmatically. I do a binary count of t bits and count the number of permutations which contain exactly n ones. because…
if a 1 represents a quake and 0 none then over t=4 months there are 3 permutations in which 2 quakes occur.
000
001
010
011 <
100
101 <
110 23 the count takes too long.
Is there a formulaic method for finding that factor? Seems the binary count might be a separate combinations problem. The result of which can be plugged into the quake occurrence formula.
Thank you and sorry for the double post.
Jerry
U should definately write a book with such a intresting method of teaching…it will help students to develop a interest in maths…
Thanx. That helped a lot. I am not joking.
Im calculating the probability of a given number oft major earthquakes occur during a given period of time. Presuming the chance of a major quake is 151/1335 in any one motnth.
p= 151/1335
t= time no of months
n= no of quakes
Prob of t quaks occuring in n months if p for one month =
p^t * (1-p)^(t-n) * number permutations of n events in t periods
I currently do it programmatic with a binary count, counting the number of permutations having n ons. But as you can imagine for n>23 too long time
what is the formula approach
cu the number of ones
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pls. help me solve this question. As a transport manager ot the TOR, you have to plan routes for your drivers. there are six deliveries to be made to customers shell, allied, esso, bosch and turrow. how many routes can be followed?