Permutation and Combination – Page 27

How-many-ways-can-the-letters-in-the-word-MATHEMATICS-be-rearranged-such-that-the-word-formed-neither-starts-nor-ends-with-a-vowel-and-any-four-consecutive-letters-must-contain-at-least-a-vowel-

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Question Number 110729 by mr W last updated on 30/Aug/20 $How many ways can the letters in$ $the word MATHEMATICS$ $be rearranged such that the word$ $formed neither starts nor ends with a$ $vowel, and any four consecutive$ $letters must contain at least a vowel ?$ Answered by…

A-blind-man-is-to-place-6-letters-into-6-pigeon-holes-how-many-ways-can-atleast-5-letters-be-wrongly-placed-Note-that-only-one-letter-must-be-in-a-pigeon-hole-

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Question Number 110524 by Aina Samuel Temidayo last updated on 29/Aug/20 $A blind man is to place 6 letters into 6$ $pigeon holes, how many ways can$ $atleast 5 letters be wrongly placed ?$ $(Note that only one letter must be in a$ $pigeon hole) .$ Answered by mr…

In-the-square-PQRS-K-is-the-midpoint-of-PQ-L-is-the-midpoint-of-QR-M-is-the-midpoint-RS-N-is-the-midpoint-of-SP-and-O-is-the-midpoint-of-KM-A-line-segment-is-drawn-from-each-pair-of-points-from-

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Question Number 110523 by Aina Samuel Temidayo last updated on 29/Aug/20 $In the square PQRS, K is the$ $midpoint of PQ, L is the midpoint of$ $QR, M is the midpoint RS, N is the$ $midpoint of SP and O is the midpoint$ $of KM . A line segment is drawn from$ $$\mathrm{each}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{points}\:\mathrm{from} \…

How-many-ways-can-the-letters-in-the-word-MATHEMATICS-be-rearranged-such-that-the-word-formed-either-starts-or-ends-with-a-vowel-and-any-three-consecutive-letters-must-contain-a-vowel-

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Question Number 110498 by Aina Samuel Temidayo last updated on 29/Aug/20 $How many ways can the letters in the$ $word MATHEMATICS be$ $rearranged such that the word formed$ $either starts or ends with a vowel, and$ $any three consecutive letters must$ $contain a vowel ?$ Commented…

in-how-many-ways-can-you-put-40-identical-balls-into-20-identical-boxes-such-that-each-box-obtains-at-least-one-ball-and-at-most-5-balls-

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Question Number 175567 by mr W last updated on 02/Sep/22 $i n h o w m a n y w a y s c a n y o u p u t 40$ $i d e n t i c a l b a l l s i n t o 20 i d e n t i c a l b o x e s$ $s u c h t h a t e a c h b o x o b t a i n s a t l e a s t o n e$ $b a l l a n d a t m o s t 5 b a l l s ?$ Terms of Service Privacy Policy Contact:…

How-many-are-the-permutations-of-1-a-little-rubik-s-cube-with-4-squares-by-side-2-a-classical-one-with-9-squares-by-side-

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Question Number 109488 by bubugne last updated on 24/Aug/20 $H o w m a n y a r e t h e p e r m u t a t i o n s o f$ $1 - a l i t t l e {r u b i k}^{'} s c u b e w i t h 4 s q u a r e s b y s i d e$ $2 - a c l a s s i c a l o n e w i t h 9 s q u a r e s b y s i d e$ Terms of Service Privacy Policy Contact: info@tinkutara.com

SUCCESSFULLY-How-many-different-words-can-you-form-using-these-letters-so-that-no-two-same-letters-are-adjacent-

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Question Number 109387 by mr W last updated on 19/Sep/20 $SUCCESSFULLY$ $H o w m a n y d i f f e r e n t w o r d s c a n y o u$ $f o r m u s i n g t h e s e l e t t e r s s o t h a t n o$ $t w o s a m e l e t t e r s a r e a d j a c e n t ?$ Commented by bobhans last updated on…

Question-43824

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Question Number 43824 by Necxx last updated on 15/Sep/18 Answered by MJS last updated on 16/Sep/18 $(\frac{1}{2})! = \frac{\sqrt{π}}{2}$ $(\frac{1}{2})! + (\frac{1}{2})! = \sqrt{π}$ $(A) 1! = 1$ $(B) \sin 45 ° = \frac{\sqrt{2}}{2}$ $$\left(\mathrm{C}\right)\:\:\sqrt{\mathrm{arccos}\:−\mathrm{1}}=\sqrt{\pi}…

Find-the-number-of-ways-a-committee-of-4-people-can-be-chosen-from-a-group-of-5-men-and-7-women-when-it-contains-people-of-both-sexes-and-there-are-at-least-as-many-women-as-men-

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Question Number 174855 by byaw last updated on 12/Aug/22 $Find the number of ways$ $a committee of 4 people can be$ $chosen from a group of 5 men$ $and 7 women when it contains$ $people of both sexes and$ $there are at least as many$ $women as men .$ Answered…

An-unfair-coin-with-the-probability-of-getting-head-in-one-toss-1-5-If-coin-tosses-n-times-the-probability-of-getting-2-heads-is-equal-to-the-probability-of-getting-3-heads-Find-n-

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Question Number 43659 by Joel578 last updated on 13/Sep/18 $An unfair coin with the probability of getting$ $head in one toss = \frac{1}{5} .$ $If coin tosses n times, the probability of$ $getting 2 heads is equal to the probability of$ $getting 3 heads . Find n$ Commented by math1967 last updated…

Category: Permutation and Combination